Question

A boy standing on a long raiboard car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of 1 $m/s^2$ and the projection velocity in the vertical direction in 9.8 m/s. How far behind the boy will the ball fail on the car ?

Answer 1

Le the velocity of car be u, when the ball is thrown.

Intial velocity of car = Horizontal velocity of ball

Distance travelled by ball B,

$S_b$ = ut (in horizontal direction)

Extra distance travelled by car

= $S_c-S_b$

= $\frac{1}{2}a{t}^2$

Ball can be considered as a projective having

$\theta ={90}^0$

t = $\frac{2usin\theta }{g}$

= $\frac{2\times 9.8}{9.8}=2sec$

$\therefore S_c-S_b=\frac{1}{2}a{t}^2=\frac{1}{2}(1{)}^2=2m$

Hence, theball will drop 2 m behind the boy.