A boy standing on a long railroad car throuws a ball straight upwards. The car is moving on the horizontal road with an acceleration of `1m/s^2` and the projectioon velocity inte vertical direction is 9.8 m/s. How far behind the boy will the ball fall on the car?

Explanation

Let the velocity of car be u, whne the ball is thrown.
Initial velocity of car = Horizontal velocity of ball.
Distnce travelled by ball B,
`S_b= ut` (in horizontal direction)
Extra distance travelled by car
`=S_c-S_b`
`=1/2 at^2`

Ball can be considered asd a projectile having
`theta= 90^0`
`t= (2u sin theta)/6
`=(2xx9.8)/9.8= 2 sec`
`:. S_c-S_b= 1/2 at^2 = 1/2 (1) 2^2= 2m`
Hence, the ball will drop 2 m behind the boy.