Question

A person is standing on a truck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8 m. Find the speed and the angle of projection (a) as seen from the truck, (b) as seen from the road.Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

Solution:
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Velocity=14.7 m/s 
Time taken by the truck to cover 58.8=t=S/v=58.8/14.7 =4sec
Since ball returns back to truck its angle of projection is vertically upwards with respect to truck.
u=speed of projection.
t=4s
h=0
h=4u-1/2 x 9.8x16
0=4u-78.4
4u=78.4
u=78.4/4=19.6 m/s


b) As seen from the road the speed of 
  ball will be resultant of 2 speeds vertical
speed =u=19.6 m/s
Horizontal speed given by truck=14.7m/s
⇒√u²+v²=√19.6²+14.7²=24.5 m/s
Thus angle seen from road:
Tan⁻¹[19.6/14.7] 
=Tan⁻¹1.33
=Tan⁻¹[4/3]
=53°with horizontal.

Answer 2

Hey mate!

Here's your answer!!

Kindly refer to the given attachment above.

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