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 speed in the vertical direction in $9.8m/s$. How far the ball will fall behind the boy? $(\text{Take}g=9.8m/s^2)$

The correct option is B $2m$
Let the initial velocity of car be $u$.
At the time of release of the ball, the horizontal velocity of ball $=u$
Time of flight of the ball is given by
$T=\frac{2u_y}{g}=2(\because u_y=9.8m/s)$
Where $u_y=$ component of velocity in vertical direction
Using second equation of motion, we get
(Distance travelled by the car in time $t=2s$)
$x_c=u\times 2+\frac{1}{2}\times 1\times 2^2=2u+2$
Distance by travelled by the ball in horizontal direction $=x_b=u\times 2$
Distance by which the ball falls behind the boy $=x_c-x_b=2u+2-2u=2m$