A man is riding on a flat car travelling with a constant speed of `10m//s`.He wishes to throw a ball through a stationary hoop `15 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 m//s` w.r.t himself.
How many seconds after he release the ball will it pass through the hoop?

Options

A. `1 sec`
B. `2 sec`
C. `3 sec`
D. `4 sec`

Answer

B

Explanation

Two important aspects to be noticed in this problem are: (1) Velocity of projection of ball is relative to man in motion
(2) Ball clears the hoop when it is at the topmost point
`vecV_(ball,man)=vecV_(ball)-vecV_(man),vecV_(ball)=vecV_(ball,man)+vecV_(man)`
Now we apply the above relation to `x`-as well as `y`-component of velocity.If ball is projected with velocity `v_(0)` and angle `theta`, then `x`-component of
`vecV_(ball)=(v_(0) cos theta+10)m//s` `y`-component of `vecV_(ball)=(v_(0) sin theta)m//s`.Since vertical component of ball's velocity is unaffected by horizontal motion of car, we can use formula for time of flight.
i.e., `(12.5 sin theta)^(2)/(2g)=5m, or sin^(2)theta=(5xx(2xx10))/(12.5xx12.5)`
or `sin theta=4/5 and cos theta=3/5`
and `v_(0)sin theta=(12.5)xx(4/5)=10m//s`
Time taken to reach maximum height `=(2v_(0)sin theta)/g`
`=(2xx10)/10=2` seconds
Horizontal distance of loop from point of projection
`=(12.5 cos theta+10)xx1=17.5 m`