Motion Under Gravity - PROJECTILE MOTION

Q1. The speed of a projectile when it is at its greatest height is √2/5 times its speed at half the maximum height. Find out the angle of projection.

Solution :-

Let θ be the angle of projection and u its initial speed. Then maximum height will be,

H = u2sin2θ/2g

So, gH = u2sin2θ/2

Now, vH = u cosθ

Or, vH2 = u2 cos2θ                  …... (1)

vH/22 = u2-2g(H/2) = u2-gH

vH/22= u2-(u2sin2θ/2)             …...(2)

Now it is given that,

vH = [√2/5] vH/2

Or, vH2 = (2/5) vH/22

Substituting the values from equations (1) and (2), we get,

(u2 cos2θ) = 2/5 [u2-(u2sin2θ/2)]

Or, 5cos2θ = 2[1-(sin2θ/2)]

Or, 5(1-sin2θ) = 2-sin2θ

Or, sin2θ = ¾

Or, sinθ = (√3)/2, Or, θ = 60o

Thus from the above observation, we conclude that, the angle of projection would be 60o.

Q2:-

A gun moving at a speed of 30m/sec fires at an angle 30o with a velocity 150m/s relative to the gun. Find the distance between the gun and the projectile when projectile hits the ground. (g = 10 m/sec)

Solution :-

Vertical component of velocity = 150 sin 30o  = 75 m/sec

Horizontal component of velocity relative to gun = 150 cos 30o = 75√3 m/sec

Horizontal component of velocity relative to ground = 75√3 + 30 ≈ 160m/sec

Time of flight   = (275)/g = 15 s

Range of projectile = 160 × 15 = 2400 m

Distance moved by the gun and projectile = 2400 - 450 = 1950 m.

Horizontal Projection:-



Consider a particle projected horizontally with a velocity  from a point O as shown in side figure.

Assuming the point of projection O as the origin of coordinates and horizontal direction as the X-axis and vertical direction as Y-axis. Let P (x, y) be the position of the particle after t seconds.

So, x = horizontal distance covered in time t = ut.         ............... (1)

y = vertical distance covered in time t = ½gt2               ............... (2)

Eliminate t from equations (1) and (2) then we get,

y = (1/2)(g/u2) (x2)

This is the equation of parabola passing through the origin, with its vertex at the origin O. Hence the trajectory is a parabola.

Q 3:-



A stone is thrown at a speed of 19.6 m/sec at an angle 30o above the horizontal from a tower of height 490 meter. Find the time during which the stone will be in air. Also find the distance from the foot of the tower to the point where stone hits the ground?

Solution :-

Let us consider the motion of stone in the horizontal and vertical directions separately.

(i) Vertical motion (downward direction negative) :

Initial vertical velocity y = 19.6 sin 30o

Acceleration a = g = -9.8 m/s2

Vertical distance covered = h = 490 m

Using, h = ut +  1/2gt2

We have, 490 = - 9.8t + (1/2) 9.8t2

100 = - 2t + t2           or     t2 - 2t - 100 = 0



t = 11.05 sec

(ii) Horizontal motion:

Initial horizontal velocity y = 19.6 sin 30o = 9.8 m/s

Hence distance from the foot of tower to the point where stone hits the ground

= Horizontal component × time of flight

= 19.6 cos 30o × 11.05 = 188 m                         

Projectile Motion on an inclined plane:-

Let the particle strike the plane at A so that OA is the range of the projectile on inclined plane. This initial velocity  can be resolved into two components:

 (i) u cos (α - β) along the plane

(ii) u sin (α - β) perpendicular to the plane.

The acceleration due to gravity g can be resolved into two components:

(i) g sin β parallel to the plane

(ii) g cos β perpendicular to the plane.

Time of Flight:-

Let t be the time taken by the particle to go from A to B. In this time the displacement of the projectile to the plane is zero.

Hence, 0 = u sin (α-β) t - ½g β t2

=> t = 2u sin(α-β)/gcosβ

Range:-

During time of flight, the horizontal velocity u cos α remains constant.

Hence, horizontal distance,

OB = (ucosα) t = 2u2sin(α-β)cosα/gcosβ

Now, OA = OB/cosβ = 2u2sin(α-β)cosα/gcosβ

The greatest distance of the projectile from the inclined plane is u2sin2 (α-β)/2gcosβ .

Problem 4:-



A Particle is projected with a velocity 39.2 m/sec at an angle of 30o to an inclined plane (inclined at an angle of 45o to the horizontal). Find the range on the incline (a) when it is projected upward (b) when it is projected downward.

Solution :-

Time of flight will be same in both cases because the acceleration perpendicular to the plane is same. Therefore,

0 = 39.2 sin 30o t - (½) g cos 45o t2

Or, t = (2×39.2 sin 30)/(g cos 45) = 4√2  sec

(a) Range upward

= 39.2 cos 30o t - (½) g sin 30o t2

= 39.2 × √3/2 × 4√2- (1/2) × 9.8 × (1/2) × (4√2)2  = 113.7m 

(b) Range downward

= 39.2 cos 30o × t + (½) g sin 30o t2

= 39.2 × √3/2 × 4√2  + (1/2) × 9.8 × (1/2) × (4√2)2  = 270.5m  Ans.

Motion down the plane:-



Let the particle be thrown at a velocity v0 at angle ‘α’ with the horizontal as shown in figure. 

v0  sin (α+β)T- 1/2 gcosβT2=0 [for y'=0]

 => T = (2v0 sin(α+β))/gcosβ 

R = v0 cos(α+ β)T+ 1/2 g sin βT2= (v02)/g  [(sin(2α+β)+sinβ)/(1-sin2β)] 

Since α is the variable and maximum value of sin function is 1, therefore for R to be maximum, sin (2α+β)=1 

and Rmax (v02)/g [(1+sinβ)/(1-sin2β)]= (v02)/(g(1-sinβ)) down the plane.