A man is standing on a rail road car travelling with a constant speed of `v = 10 ms^-1` (Fig . 5.34). He wishes to throw a ball through a stationary hoop `5 m` above the height of his hands in such a manner that the ball will move horizontally as it passes through the hoop. He throws the ball with a speed of `12.5 ms^-1` w.r.t. himself.
(a) What must be the vertical component of the initial velocity of the ball ?
(b) How many seconds after he releases the ball will it pass through the hoop ?
( c) At what horizontal distance in front of the loop must he release the ball ?
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Explanation

The important aspects to be noticed in this problem are :
The velocity of the projection of ball is relative to man in motion.
The ball clears the hoop when it is at the topmost point.
`vec v_(ball,man) = vec v_(ball) - vec v_(man)`
`vec v_(ball) = vec v_(ball,man) + vec v_(man)`
(a) Now we apply the above relation to x - as well as y - component of velocity. If the ball is projected with velocity `v_0 and angle theta`, then
x - component of `vec v_(ball) = (v_0 cos theta + 10) ms^-1`
y - component of `vec v_(ball) = (v_0 sin theta) m s^-1`
(b) Since the vertical component of the ball's velocity is unaffected by the horizontal motion of car, we can use the formula for the time of flight.
`((12.5 sin theta)^2)/(2 g) = 5 m`
`sin^2 theta = (5 xx (2 xx 10))/(12.5 xx 12.5)`
`sin theta = (4)/(5) and cos theta = (3)/(5)`
`v_0 sin theta = (12.5) xx ((4)/(5)) = 10 m s^-1`
Time taken to reach the maximum height,
`(2 v_0 sin theta)/(g) = (2 xx 10)/(10) = 2 s`
( c) The horizontal distance of loop from the point of projection `= (12.5 cos theta + 10) xx 1 = 17.5 m`.