Question

A boy standing on an open car throws a ball vertically upwards with a velocity of $9.8m/s$, while moving horizontally with uniform acceleration of $1m/s^2$ starting from rest. The ball will fall behind the boy on the car at a distance of

• A. $1m$
• B. $2m$
• C. $3m$
• D. $4m$

Answer 1

The correct option is B $2m$

Initial velocity of the ball
$u=9.8m/s$
$v=0$ (at the top)
So, $v=u-gt$
$0=9.8-9.8t$
$t=1$
So, to reach the height point the ball takes $1$ sec.
So, time taken by the ball to come back
$=2t=2sec$
in $2$ sec, the car advances
$S=ut+\frac{1}{2}+t^2$
$=0+\frac{1}{2}1\times 4=2m$
So the ball will fall $2$ m behind the car.