Motion In A Plane

Vector:The quantity which has magnitude as well as direction E.g. Velocity, Acceleration etc.

Scalar: The quantity which has magnitude only E.g. mass , volume, speed etc

The two scalar quantities are work and current because they are defined by magnitude only.

NOTE: A scalar quantity is a quantity that is described by a magnitude or numerical value only.

Work has magnitude only. Although current has direction also, it is a scalar quantity because it follows general laws of algebra.

The vector quantity is impulse.

NOTE: Impulse is defined as the change in momentum. It gives the effect of a force over a period of time.

J = Δp = FΔt

Where, J is the impulse

P is the momentum

F is the force

t is time

Here, p and F are vectors so that J is also a vector.

(a) Addition of two scalars is meaningful since both follow general algebraic laws.

(b) Addition of a scalar to a vector of the same dimensions is not meaningful since the scalar follows general algebraic laws but the vector does not.

(c) Multiplying any vector by a scalar is meaningful as the magnitude of the vector is multiplied by the scalar. For example, force which is a vector when multiplied by the scale time gives the vector impulse.

(d) Multiplying any two scalars is meaningful irrespective of their dimensions. For example, speed multiplied by time gives the distance travelled. Here, speed and time are scalars which give distance which is also a scalar.

(e) Addition of any two vectors is meaningful only if they have the same dimensions. This addition takes places according to vector algebra.

(f) Adding a component of a vector to the same vector is meaningful because both of them have the same dimensions.

NOTE: A scalar quantity is a quantity that is described by a magnitude or numerical value only. A vector quantity is a quantity that is described by both magnitude and direction.

(a) True

Explanation: The magnitude of a vector is always a scalar because magnitude is a numerical value.

(b) False

Explanation: Each component of a vector is not always a scalar. Component of a vector is the effect of the vector in a given direction. It is also a vector.

(c) False

Explanation: The total path length is not always equal to the magnitude of the displacement vector of a particle. It is only true for a motion in straight line where the distance is equal to the magnitude of displacement.

(d) True

Explanation: The average speed of a particle is either greater or equal to the magnitude of average velocity of the particle over the same interval of time. Speed is the ratio of distance travelled to the time taken. Velocity is the ratio of displacement of an object to the time taken. For a given path over a given time period, distance is always greater than or equal to the displacement of the object. Hence, speed is always greater than or equal to velocity. The equality condition occurs for a straight line.

(e) True

Explanation: Three vectors not lying in a plane can never add up to give a null vector. For the resultant to be a null vector, the three vectors should form a triangle. The three sides of a triangle necessarily lie on the same plane. Hence, the given statement is true.

Let us consider two vectors and such that = and =. Also,=. According to Parallelogram law of vector addition, = and = as shown in the figure.

From the figure,

OA=

OB=AC=

OC=

OC’=

A. In a triangle, each side is smaller than the sum of other two sides.

So, in ΔAOC,

OC < OA + AC

⇒

If both the vectors act along a straight line, then the equality condition occurs as

So, =

∴

B. In ΔAOC,

OC+AC>OA

⇒ |OC|>|OA-AC|

⇒ |OC|>|OA-OB| (∵ AC=OB are the parallel sides of the parallelogram)

⇒

If both the vectors act along a straight line, then the equality condition occurs as

∴

C. In ΔOAC’,

OC’<OA+AC’

⇒

⇒ (∵ )

If both the vectors act along a straight line, then the equality condition occurs as

∴

D. In ΔOAC’,

OC’+AC’>OA

⇒ |OC’|>|OA-AC’|

⇒

⇒

If both the vectors act along a straight line, then the equality condition occurs as

∴ ⇒

NOTE: Parallelogram Law of Vector Addition states that when two vectors are represented by two adjacent sides of a parallelogram by direction and magnitude then the resultant of these vectors is represented in magnitude and direction by the diagonal of the parallelogram starting from the same point.

A. Incorrect

Explanation: It is not necessary that the vectors ,, and be null vectors for their sum to be a null vector. There can be other combinations as well. For example, if then

.

B. Correct

Explanation:

⇒

Taking modulus on both sides,

⇒

Hence, magnitude of () is equal to the magnitude of ().

C. Correct

Explanation:

⇒

Taking modulus on both sides,

⇒

Hence, the magnitude of is always equal to the magnitude of and can never be greater than that.

D. Correct

Explanation: For the sum to be a null vector, the vectors , and must be three sides of a triangle according to triangle law of vector addition. The three sides of a triangle lie on the same plane. Hence, the vectors , and must be coplanar. But if and are collinear, then must lie on the same line and in opposite direction in order to cancel out in the sum .

NOTE: Triangle law of vector addition states that when two vectors are represented by two sides of a triangle in magnitude and direction taken in same order then third side of the third side of that triangle represents in magnitude and direction the resultant of the vectors.

Displacement of an object is given by the change in its position. Here, the initial position is P and the final position is Q for all three girls. So, the displacement is same for all the three girls and its magnitude is equal to the diameter of the circular ice ground.

So, Magnitude of Displacement = 2×Radius = 2×200m=400m

The magnitude of the displacement is equal to the actual path length travelled for girl B since she moved along the diameter.

(a) Displacement is given by the change in position. In the given case, the initial position is also the final position i.e., there is no change in position. So, net displacement is zero.

(b) Average velocity is the ratio of net displacement to the time taken.

i.e.,

Since net displacement is zero, average velocity is also zero.

(c) Average speed is the ratio of total path length to the time taken.

i.e.,

Here, total path length = OP+PQ+QO

⇒ total path length = r + + r

(∵ Circumference of a circle = 2πr)

⇒ total path length = r×(2+)

⇒ total path length = 1km× (2+1.57)

So, total path length = 3.57km

Also, time taken = 10min (given)

Hence, average speed =

⇒ average speed =

⇒ average speed = 21.42 km/hr

The entire motion of the motorist can be shown using the following diagram:

Let A be the starting point of the motorist.

At the third turn:

At the third turn, his position is D.

Magnitude of Displacement = AD

⇒ Magnitude of Displacement = AO + OD

⇒ Magnitude of Displacement = 500m + 500m

⇒ Magnitude of Displacement = 1000m

Total path length = AB + BC + CD

⇒ Total path length = 500m + 500m + 500m

⇒ Total path length = 1500m

At the sixth turn:

At the sixth turn, the position of the motorist is A.

So, the final position is the initial position i.e., there is no change in position.

Hence, magnitude of displacement = 0

Total path length = AB + BC + CD + DE + EF + FA

⇒ Total path length = 500m+500m+500m+500m+500m+500m

⇒ Total path length = 3000m

At the eighth turn:

At the eighth turn, the position of the motorist is C.

Magnitude of displacement = AC

⇒ Magnitude of displacement =

⇒Magnitude of displacement =

⇒ Magnitude of displacement =

⇒ Magnitude of displacement =

∴ Magnitude of displacement = 866.025m

Total path length = AB + BC

⇒ Total path length = 500m + 500m

∴ Total path length = 1000m

NOTE: Total path length or distance travelled is the length of the actual path taken by an object whereas displacement is the change in its position i.e., the shortest distance between the final position and the initial position. Path length is a scalar quantity whereas displacement is a vector quantity.

Given:

Shortest distance from station to hotel(d₁) = displacement= 10km

Circuitous path distance from station to hotel(d₂) = 23km

Time taken to reach the hotel (t) = 28min

= 28/60 hr

= 0.467 hr

(a)

=

= 49.25 km/hr

Explanation: When speed of the body is calculated, the distance travelled by it other than the shortest distance is considered.

Average speed of the taxi is 49.25 km/hr

(b)

=

= 21.413 km/hr

Explanation: Velocity of a body is being calculated, the shortest/ straight line distance travelled by it is considered

Figure showing the directions of velocities of rain and cycle

Explanation: Direction in which the umbrella should be held is opposite to the direction of velocity of rain with respect to cycle.

V_r = Velocity of rain

V_c = Velocity of cycle

Velocity of rain w.r.t the cycle = V_rc = V_r-V_c

= 30 m/s + (-10)m/s

= 20 m/s

= 0.33

θ = tan^-1(0.33)

= 18.26˚

Hence, the woman must hold the umbrella towards the south at an angle of 18.26˚ to the vertical direction.

Figure showing the velocities of swimmer and river

Given:

Speed of the man = V_m = 4km/hr in still water

Width of the river = d = 1km

Speed of the river = V_r = 3 km/hr

Explanation: The time taken by the man to cross the river is the ratio of width of the river and speed of the man. The velocity displaces the man down the river.

= = 0.25hr

= 15 min.

Distance covered due to the flow of river = l= V_r×t

= 3 km/hr × hr

= 3/4 km

= 750m

Time taken by man to cross the river is 15 min and distance covered due to flow of water is 750m down the river.

Figure showing velocity of wind and the boat

Given:

Velocity of wind = V_w = 72 km/hr along N-E direction

Velocity of boat = V_b = 51 km/hr along N direction

Explanation: To find out the direction of flag on the mast of the boat, we need to find out the relative velocity of wind w.r.t the boat.

Relative velocity of wind w.r.t the boat = V_wb = V_w - V_b

= 72 - V_b

Angle between V_w and -V_b = 135˚

Using parallelogram law of vectors, let V_wb be the resultant vector making an angle β with V_w

= 1.0034

β = 45.099˚

= 4.1˚ w.r.t N-E direction

i.e. 45.1˚-45˚ = 0.1˚ w.r.t East

The direction of flag on the mast of the boat is 0.1˚ w.r.t East.

Given:

Height of the hall= H = 25m

Velocity of the ball = Velocity of the projectile = v = 40 m/s

By applying the concept of projectile motion, we know that

ϕ= Angle that the projectile makes with the horizontal

g= Acceleration due to gravity= 9.8 m/s²

Sin ϕ = 0.5533

⇒ ϕ = sin^-1(0.5533)

⇒ ϕ = 33.6˚

R = Range of the projectile (or) horizontal distance travelled by the projectile.

= m

= 150.5m

Therefore, the ball can cover a maximum distance of 150.5m without hitting the ceiling of the wall.

Given:

Maximum horizontal distance travelled by the ball=Range =R= 100m

Explanation: We know that range is maximum if the angle of projection is 45˚.

Figure showing projectile motion

(ϕ = 45˚)

u = 31.304 m/s

Height to which the ball is thrown = H

Since the final velocity of the ball is zero i.e. v= 0

From the relation v²-u²=2as

⇒ 0 - 31.304² = 2×9.8×H (s=H)

⇒ H= 50m

The cricketer can throw the ball to a height of 50m above the ground.

Given:

Length of the string = l= 80cm = 0.8m

Horizontal circular motion of a stone tied to a string

= H

=(2××) rad/s

ω = rad/s

Explanation: The object is in circular motion. Therefore, magnitude of acceleration produced in the stone is equal to the magnitude of centripetal acceleration.

Centripetal acceleration = a_c = ω²r

= m/s²

= 9.91 m/s²

Acceleration is towards the centre of the circle along the radius.

Centripetal acceleration is 9.91m/s² towards the centre of the circle along the radius.

Given:

Radius of the horizontal loop = r = 1 km = 1000m

Speed of the aircraft = v = 900 km/hr

= 250 m/s

= 62.5 m/s²

Acceleration due to gravity = g= 9.8 m/s²

a_c/g = 62.5/9.8 = 6.38

Ratio of centripetal acceleration and acceleration due to gravity is 6.38.

A) False, because the net acceleration of a particle in circular motion along the radius can be both towards the centre of the circle and away from it. It is towards the centre of the circle only in case of uniform circular motion.

B) True, because while leaving the circular path, the particle moves tangentially to the circular path.

C) True, the acceleration of particle in a uniform circular motion is a vector directed towards the centre of the circular path. The sum of these vectors over one complete cycle is a null vector.

_{(A) velocity is rate of change of displacement w.r.t time}

_{Velocity =}

Acceleration

= m/s²

(B) At t= 2.0 s

Velocity = v = m/s

= m/s

Magnitude of velocity = 8.544 m/s

tanθ = =-0.375

θ = -20.55˚

Acceleration = a = m/s²

Magnitude of acceleration = 4

tanθ = 0

θ = 0˚

Velocity is 8.544 m/s at an angle of -20.5˚ with the horizontal. Acceleration is 4 m/s² in the negative X direction.

Here the velocity and acceleration of particle are given in vector form and are unit vectors along X axis and Y axis respectively

i.e. they are vectors with magnitude unity and are representing X direction, and Y direction, each vector in x-y plane can be resolved into two components one along with x-direction and other along the y-direction, the vector quantity is the linear combination of both the components

the initial velocity of the particle is

i.e its magnitude is 10.0 m/s in y direction

since coefficient of is zero so magnitude in x direction is zero

the acceleration of the particle is

i.e. the magnitude in x-direction is 8.0 m/s and magnitude along y-direction is 2.0 m/s

(a) since the particle started from the origin, so its displacement in x-direction will give its new x coordinate and displacement in y-direction will give its new y coordinate, so the displacement of the particle in the x-direction is

S_x = 16m

i.e. when x co-ordinate of a particle is 16m when time t = 2s
Now putting the value T=2 sec

(b) Speed is the magnitude of the velocity of the particle so we will first find the component of velocity in x and y-direction using the equation of motion.
Now using the final velocity equation, we get

The speed of the particle

So the speed of the particle is 21.26 m/s

Any vector quantity in x-y plane is represented as linear combination of two unit vectors one along x direction and other along y direction namely and , coefficient of and represent magnitude of vector quantity in x and y direction respectively

for any vector

of p_x and p_y represent magnitude of vector quantity in x and y direction respectively

the magnitude is given by

And angle with x axis is given by

Here let the vectors be

and

Now considering

X component a_x = 1 unit

Y component a_y = 1 unit

So magnitude of vector is

Or

Now for direction angle with x axis

Or

So angle made with x axis is

Now considering

X component b_x = 1 unit

Y component b_y = -1 unit

So magnitude of vector is

Or

Now for direction angle with x axis

Or

So angle made with x axis is

As shown Graphically in the figure

Now component of one vector in the direction of other vector is the magnitude of vector in direction of other

Now let us say we have to find component of in the direction of let us angle between them is 𝜽 now any vector can be resolved into rectangular components parallel to another vector so component of in the direction of is qCos𝜽

as can be shown in figure

Now to find q.cos𝜽

We know dot product or scalar product of two vectors will be

Where p and q are magnitudes of vector and respectively

And 𝜽 is angle between them

Solving we get

Or magnitude of along is

We are given

And we have to find its component along

and
Now component of along is where is magnitude of

We know,
magnitude of = units

So, component of along is units units

Now , the angle between the x and y component of vector is


Angle between the vector and vectors

The component along the vector has a magnitude of

Angle between the and vector

The magnitude of component along the vector

The magnitude of component of and vector is

(The ‘average’ stands for average of the quantity over the time interval t₁ to t₂)

(a) False

If the velocity of particle is v (t₁) at time t = t₁ and velocity is v (t₂) at time interval t = t₂ the average velocity of particle is denoted by

v_average = (1/2) (v (t₁) + v (t₂))

Only in case of uniform motion that is when acceleration of the particle is constant or the rate of change of velocity is fixed, in case of arbitrary motion when acceleration may or may not be constant the above relation is incorrect

(b) True

We know the average velocity for any type of motion is given by v_average = total displacent / total time

Now r(t₁) and r(t₂) are the position vector of particle at time interval t = t₁ and t = t₂ respectively so r(t₂) - r(t₁) is the change in position of particle or the displacement of the particle and ( t₂ – t₁) is the time taken for that displacement so [r(t₂) - r(t₁) ] /( t₂ – t₁) denotes the average velocity of the particle v_average

(c) False

Here v (0) denotes the velocity of the particle at time interval, t = 0 i.e. it denotes initial velocity of the particle v(t) is the velocity of particle at any time interval t seconds and a is the acceleration, so it is the first equation of motion v = u + at

Where v is the final velocity after time t, u is the initial velocity and a is the acceleration, but this equation is valid only in case of uniform motion i.e. when acceleration of particle is constant but in case of arbitrary motion it may or may not be constant so above equation is not always valid.

(d) False

Here r (0) denotes the position of particle at time interval t = 0 , r (t) is the position of particle at time interval t , a is the acceleration of particle

r (t) = r (0) + v (0) t + (1/2) at² can be rearranged as

r (t) - r (0) = v (0) t + (1/2) at² so r (t) - r (0) is the change in position of particle or displacement v (0) is the velocity of particle at time interval t = 0 or initial velocity of the particle i.e. this equation represents the equation of motion

S = ut + ( �)at² where S is the displacement of particle u is the initial velocity of the particle a is acceleration and t is time but the above equation is valid only in case of uniform motion i.e. when acceleration of particle is constant but in case of arbitrary motion it may or may not be constant so above equation is not always valid.

(e) True

We know the average acceleration for any type of motion is given by a_average = change in velocity / total time

Now v(t₁) and v(t₂) are the velocity of particle at time interval

t = t₁ and t = t₂ respectively so v(t₂) - v(t₁) is the change in velocity of particle and ( t₂ – t₁) is the time taken for velocity to change so [v(t₂) - v(t₁) ] /( t₂ – t₁) denotes the average acceleration of the particle a_average

A. False

A scalar quantity is one which only has magnitude and no direction for e.g. Speed, Pressure, Energy, Temperature etc. they need not always be conserved in a pressure for e.g. In most thermodynamic process pressure, Temperature of the system varies, in inelastic collisions Energy of particles are not conserved etc. so we cannot say scalar quantity is always conserved in a process

B. False

Scalar quantities do not have a direction but they can take both positive and negative values and the sign does not depict direction for eg. Temperature is a scalar quantity and can have negative values but its negative sign does not depict direction because it does not have a direction it’s just a value less than reference 0 value

C. False

Scalar quantities, just like vector quantities have units hence dimensions as they have magnitude which is measured with appropriate unit and hence have a dimension for e.g. Distance is a scalar quantity and is measured in metres or its unit is metre(m) so its dimension is L ,Speed is a scalar quantity and its unit is m/s so its dimension is LT^-1 likewise scalar quantities have dimensions.

D. False

Scalar quantities can vary from point to point in space and values need not be constant in space just like vector quantities for eg. Potential energy is a scalar quantity, and potential energy is body possessed by the body due to virtue of its position and as the position changes the potential energy certainly changes like gravitational potential energy of an object depends on the height from ground i.e. the quantity vary with change in position and hence vary from one point to another in space

E. True

Scalar quantity does not have a direction, it only has magnitude and if orientation of axes is changed only direction of observation changes and magnitude remain same but since scalar quantity does not have a direction, direction of observation is meaningless and as magnitude is constant to there is no change at all.

now let the aircraft be initially at a point A situated 3400m above the ground, it moves with a speed v along straight line parallel to ground and reach a point B, now let us assume a recording station is located on ground at c which is in line with midway of the journey D

Suppose the plane covered a distance S and subtended an angle of 30^o on recording station as it moved from A to B

The situation has been shown in the figure

We have to find speed of the plane; the plane is moving with constant speed so we will use the relation

S = v × t or v = S/t

Where S is the distance covered by plane moving with a speed v in time t

Here we are given time t = 10.0 s

So we need to find distance covered by plane S

Now we can see distance covered = AB = AD + BD

Since D is the midpoint of AB

AD = BD or distance AB = 2 × AD

Now in Triangle ACD using trigonometry we will find AD

now

Assuming D is the midpoint of AB and C is equidistant from A and B we have

Or

AD = CD × tan15^o

CD = 3400m

tan15^o= 0.2679

so we get

AD = 3400m × 0.2679

= 910.86m

So total distance covered by the plane

S = AB = 2 × 910.86m = 182.17m

So the Value of S = 182.17m and t = 10 s in the equation

v = s/t ,we get

speed of the plane

so plane is flying at a speed of 18.2 m/s

A vector has magnitude and direction but in general it does not have a fixed location of space because a vector can be translated parallel to itself or we can say if a vector is moved parallel to itself keeping its direction and magnitude same then the vector is same or there is no effect on vector as can be shown in figure

Now here there are three vectors at A,B,C all have same length so have same magnitude, have same direction towards positive x axis and are thus parallel to each other so all the three vectors are same so there is no effect of location i.e. position is not fixed but in case of position vector position of each point is different and position vector denotes position in terms of co-ordinates of x and y so position vector have a fixed location and are also directed from origin so two position vectors cannot be parallel if they are denoting different positions so position vector have definite position in space but in general all vectors does not have a specific position in space

Yes , vector can certainly vary with time and many vector quantities are just rate of variation of other quantities or vectors can be a function of time for e.g. Velocity of a particle in uniform motion is a function of time and as the time increases the velocity of particle increases or decreases depending upon the acceleration of particle i.e. velocity changes with time likewise in general vector can vary with time

Now we have two equal vectors a and b at different locations in space they necessarily need not have same physical effects though in specific cases they can have same physical effects but this is not true always for e.g. two forces of same magnitude and same direction applied on a body fixed lever can produce different turning effects as shown in figure

As can be seen both a and b are directed vertically downwards i.e. same direction and have same magnitude so both are equal but turning effect will be different due to them because their location is different in same so we conclude that equal vectors a and b at different locations in space do not necessarily have identical physical effects

A vector quantity is one which has both magnitude and direction but should also obey laws of vector addition subtraction and multiplication for e.g. Electric Current has both magnitude and direction but is not a vector it is added algebraically not vectorially so it is a scalar quantity suppose current I₁ and I₂ are flowing as shown in figure both have same magnitude the net current is I = I₁ + I₂ = 2 I₁ = 2I₂

If if we would have added them using vector laws, both are equal in magnitude and opposite in direction so resultant would have been zero which is incorrect so current is not vector

likewise, all quantities having both magnitude and direction are not vectors

Now suppose a body is rotating about a fixed axis and has rotated by an angle 𝜽 now we have both direction of axis of rotation and the angle of rotation, but this rotation is not a vector because direction of rotation cannot be specified with these two quantities suppose there is a particle at A reached B

as shown in figure

Now direction of rotation is continuously changing and the total angle of rotation 𝜽 describe the magnitude of rotation but the direction cannot be specified but if we take an very small angle of rotation i.e. at each instant direction can be specified when we are considering change in angle is very small i.e. instantaneous change in angle of an rotating body d𝜽 and they obey the law of vector algebra as well and direction can be specified as tangential direction as shown in figure

So we can see if we take very small instantaneous angle of rotation then rotation is vector but if we take large angle direction cannot be specified and hence it’s not a vector so every rotation is not vector i.e. any rotation is not vector

A vector quantity is one which has both magnitude and direction and also obeys laws of vector algebra

(a) now considering a length of wire bent into loop it has a magnitude that is its length but no specific fixed direction the length at each instant can have a unique direction but no unique direction of the whole loop so it cannot be represented by a vector as shown in figure

(b) Now an Plane Area can be described as a vector since it has a magnitude that’s the area measured in m² and the whole plane area can be described by a vector whose direction shows the direction of normal to area and for a plane area normal is unique so direction of the whole plane area is unique as shown in figure

Now in general a plane area with surface area S can be described in vector form as

Where denotes the area vector S denote the magnitude of surface area and is a unit vector in the direction of normal or we can say perpendicular to the plane area

(c) Now a sphere have volume and surface area now volume also does not have a definite direction and has only a magnitude measured in m³ at each instant volume of a body or sphere will have different direction so volume of sphere cannot be associated with sphere because the whole volume cannot be represented by a unique direction now if we take area into account the normal at any every point of sphere will have a different direction directed outwards normally there is no unique direction to specify the whole area of sphere as shown in figure

So, the sphere cannot be associated with a vector because it there is no unique direction which could specify whole sphere.

Now this is the case of projectile motion, the motion of bullet will be a projectile and its path will be parabola, the motion of bullet can be divided into horizontal component and vertical component i.e. motion along x axis and motion along y axis

Now the bullet is only experiencing one force in vertically downward direction because of its weight (gravitational force of earth) due to which its accelerating along negative y axis due to acceleration due to gravity, acceleration is uniform i.e. all the three equations of motion are valid in along Y direction, now there is no force in horizontal direction so component of velocity in horizontal direction or along X axis would remain constant.

The projectile motion of bullet has been shown in the figure

Now for an particle projected at an angle 𝜽 with horizontal plane or ground at an initial velocity u the maximum displacement of the particle in horizontal plane or x direction also called the range of projectile is given by

Where R is the range of the particle, u is the initial velocity of the particle, 𝜽 is the angle of projection of the particle and g denotes acceleration due to gravity

g = 9.8 ms^-2

Now in the case of bullet we are given

Range of the particle,

R = 3 Km = 3000m

Angle of projection,

𝜽 = 30^o

Let u the initial velocity of the particle

u = ?

putting the values in the equation

we get,

Or putting

Or

Now we have to adjust the gun so that bullet can go upto 5 Km or we can say range is increased to 5 Km

We can only change the angle of projection and its initial velocity u will be same or value of u² will be same

Now let tange of projectile be R’

R’ = 5Km = 5000m

We have u² =

We have to find angle of projection,

𝜽 = ?

Putting these values in equation

We get

Or,

But sin2𝜽 = 1.44 is not possible as for any value of 𝜽 maximum value of sine function is 1 i.e. the equation is invalid so the range R’ cannot be 5 Km so bullet cannot hit a target at a distance of 5 Km from it

The maximum range of particle is when 𝜽 = 45⁰ when

sin2𝜽 = sin 90^o =1

and maximum range is

Putting value of square of initial velocity of the particle

u² =

we get Range

So the bullet can maximum hit a target 3.46 Km away

Here the plane will move forward in horizontal direction in some time t and shell will cover the same horizontal distance along with reach the altitude or height of the plane

Now this is the case of projectile motion, the motion of Shell will be a projectile and its path will be parabola, the motion of shell can be divided into horizontal component and vertical component i.e. motion along x axis and motion along y axis

Initial speed of shell is u = 600 m/s

And let it be making an angle 𝜽 with the vertical

So horizontal component of velocity

u_x = u × sin𝜽 = 600sin𝜽

and vertical component of velocity

u_y = u × cos𝜽 = 600cos𝜽

Here the velocity of fighter plane is uniform and is in horizontal direction let it be

v₁ = 720 km/h

converting it to m/s

1 Km = 1000m and 1 hour = 60 × 60 = 3600 s

So v₁ = 72 × 1000m/3600s = 720 × 5/18 = 200 m/s

Let us consider in time t sec plane moves a distance X with uniform speed 20 m/s is horizontal direction in the same time the shell moved same distance X in same time t in horizontal direction due to horizontal component of velocity and moved a distance of Y = 1.5 Km = 1500 m equal to altitude of plane in vertical direction in order to hit the plane

The situation has been shown in the figure

Now the shell is only experiencing one force in vertically downward direction because of its weight (gravitational force of earth) due to which its accelerating along negative y axis due to acceleration due to gravity, acceleration is uniform i.e. all the three equations of motion are valid in along Y direction, now there is no force in horizontal direction so component of velocity in horizontal direction or along X axis would remain constant.

Since shell and plane cover equal distance in equal time in horizontal direction with uniform speed so horizontal component of velocity of shell and speed of plane must be equal

i.e u_x = v₁

or u × sin𝜽 = 600 × sin𝜽 = 200 m/s

or sin𝜽 = 20/600 = 1/3

now so we get

so the gun must be making an angle of 19.47^o with the vertical

In order to avoid being hit by the gun the plane must fly above the the maximum vertical height shell can reach, the motion of shell is a projectile as explained above and for the maximum height of projectile we will apply equation of motion

in y direction

Where v is the final velocity, u is the initial velocity of the particle, a is the acceleration and s is the displacement of the particle

For y direction we have

Where Initial velocity of particle in vertical direction making an angle 𝜽 with the vertical

u_y = u × cos𝜽 = 600cos𝜽

since upon reaching maximum height Shell will come to rest to final velocity in vertical direction

v_y = 0 m/s

displacement of shell in y direction must be equal to height of plane let the maximum height of plane be

S_y = Y = ?

The acceleration of particle is in downward direction and is equal to acceleration due to gravity

a �_y = -g = -10 ms^-2

putting values in equation)

0² – (600ms^-1× cos𝜽)² = 2 × (-10 ms^-2) ×Y

So we have

Here cos²𝜽 = cos²(19.47^o)

We know sin²(19.5^o) = (1/3)²=1/9

Using cos²𝜽 + sin²𝜽 = 1

cos²(19.47^o) = 1 - sin²(19.5^o) = 1 – 1/9 =8/9

so putting value, we know maximum height of shell can be

So the fighter plane must fly above 16 Km from ground in order to avoid being hit by shell

Now suppose the Particle is moving along x axis in positive x direction now its speed is reducing i.e. its de-acceleration or the direction of tangential acceleration (acceleration due to increase or decrease in speed) is towards negative x axis , there is another component of acceleration radial acceleration because direction of particle is also changing as it is following a circular path now direction of radial acceleration is perpendicular to the velocity of particle i.e. along Y axis

As shown in figure

Since the speed is decreasing by 0.5 m/s every second, the rate of change of speed is the tangential acceleration so the tangential acceleration is

a_t = -0.5 ms^-2 i.e. 0.5 ms^-2 opposite to the speed of cyclist along negative X axis

now radial acceleration of a particle undergoing circular motion is given by

a_r = v²/r

where v is the magnitude of velocity or the speed of the particle,

r is the radius of the circular patch

now at the beginning of path speed of the particle is

v = 27 km/h

converting it to m/s

1 Km = 1000m and 1 hour = 60 × 60 = 3600 s

So v= 27 × 1000m/3600s = 27 × 5/18 = 7.5 m/s

Radius of the circular path

r = 80 m

so the radial acceleration is

a_r= 0.70 ms^-2 radially inwards i.e. perpendicular to velocity of cyclist along positive y axis

so the magnitude of resultant acceleration or net acceleration of the cyclist is given by

i.e.

a = 0.86 ms^-2

i.e. the net acceleration of the cyclist is 0.86 ms^-2 the direction is as shown in the figure above now let us find the angle made by the net acceleration of particle with radial acceleration 𝜽 which is given as

Where a_r is the magnitude of radial acceleration

here a_r = 0.7 ms^-2

and a_t is the magnitude of tangential acceleration

here a_t = 0.5 ms^-2

so we get

i.e. the total acceleration of cyclist is making an angle of 54.5^o with the tangential acceleration of cyclist

Suppose in an projectile motion an particle is thrown making some angle 𝜽_o with the horizontal with initial velocity v_o sow components of the initial velocity in horizontal and vertical direction are given as

v_0x = vcos and v_0y = vsin𝜽

there is no horizontal force on the particle so horizontal component of velocity will remain constant and due to acceleration due to gravity the motion in y direction will be uniformly accelerated where y component of velocity will change and all equations of motion will be valid in y direction

A. Now at any instant of projectile motion let the angle of velocity of particle with x axis or horizontal be 𝜽(t) now 𝜽(t) is varying because y component of velocity of particle is constantly changing with time so the angle of resultant velocity is also changing with time , now angle of resultant velocity of the particle or the velocity of particle with x axis at any instant is same as angle with x component or horizontal component of velocity at any instant let the y component of velocity of particle be v_y and x component of velocity of particle be v_x.

As shown in figure

So now angle of resultant velocity v at any instant with horizontal component of velocity v_x is (t)

So

Now since x component of velocity constant and equal to v_0x

So we have v_x = v_0x

Now to find y component of velocity at any instant we will aplly equation of motion

v = u + at in y direction

here v is the final velocity, u is the initial velocity t is the time and a is the acceleration of particle

for y direction velocity of particle in y direction is

v = v_y

initial velocity of particle in y direction

u = v_oy

particle is accelerating because of acceleration due to gravity is in vertically downward direction or negative y direction so

a = -g (g is acceleration due to gravity)

so y component of velocity of particle at any time t sec is

v_y = v_oy + (-g)t

or v_y = v_oy – gt

so putting values of v_y = v_oy – gt and v_x = v_0x

we have

hence angle between the velocity and x axis at any instant is given by

B. Now let the projectile be launched at initial angle 𝜽_o with the horizontal distance covered by the projectile i.e. its range be R and the maximum height achieved by projectile be h_m

as shown in the figure below

Now for an projectile the maximum horizontal distance or the range is given by the equation

Where R is the range, v_o is the in initial velocity of the particle, 𝜽_o is the angle of projection and g is the acceleration due to gravity

Now the maximum height of the projectile in vertical direction is given by the equation

Where h_m is the maximum height achieved by the particle, 𝜽_o is the angle of projection and g is the acceleration due to gravity

Now diving these equations, we get

Or

Since sin2𝜽 = 2sin𝜽cos𝜽

And cancelling out g and v_o we get

Or we have

So we get

So

i.e. the initial angle of projection for an projectile can be given as