A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of 1 ms^-2 and the projection velocity in the vertical direction is 9.8 ms^-1. How far behind the boy will the ball fall on the car? (in meters)

Let the velocity of car be u when the ball is thrown.

Initial velocity of car = Horizontal velocity of ball.

Distance travelled by ball B,

${\mathrm{S}}_{\mathrm{b}}=\mathrm{ut}\text{ (in horizontal direction) }$

Car has travelled extra distance

$={\mathrm{S}}_{\mathrm{c}}-{\mathrm{S}}_{\mathrm{b}}=ut+\frac{1}{2}{\mathrm{at}}^2-ut=\frac{1}{2}{\mathrm{at}}^2$

Ball can be considered a projectile having $\mathrm{\theta }$ = 90⁰

$\begin{array}{l}\mathrm{t}=\frac{2\mathrm{usin}\mathrm{\theta }}{\mathrm{g}}=\frac{2\times 9.8}{9.8}=2\mathrm{s}\\ {\mathrm{S}}_{\mathrm{c}}-{\mathrm{S}}_{\mathrm{b}}=\frac{1}{2}{\mathrm{at}}^2=\frac{1}{2}\left(1\right)2^2=2\mathrm{m}\end{array}$

Hence, the ball will drop 2 m behind the boy.