Question

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 $1m{s}^{-2}$ and the projection velocity in the vertical direction is $9\cdot 8m{s}^{-2}$. How far behind the boy will the ball fall on the car?

Answer 1

Given $a=1$

Let Velocity of car$=u$ ( when ball is thrown)

$\Rightarrow Initialvelocityofcar=horizontalvelocityofball$

Distance travelled by ball B, $s_b=ut$

$\Rightarrow$Car will travel extra distance $=s_c-s_b=\frac{1}{2}a{t}^2$

Consider the ball as a projectile having θ=90°

$\Rightarrow t=\frac{2u\sin \theta }{g}$

$\Rightarrow t=\frac{2\times 9.8\sin {90}^0}{9.8}$

$\Rightarrow t=2$sec

Therefore, $s_c-s_b=\frac{1}{2}a{t}^2=\frac{1}{2}{2}^2=2m$

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