A long railroad car is moving along a horizontal road with constant acceleration a. A boy standing on the car throws a ball vertically upward with velocity 9.8 m/s. If the ball lands on the car at a distance of 2 m from the boy, then the value of a (in m/s²) is (take $\left.g=9.8\mathrm{m}/{\mathrm{s}}^2\right)$

Let us solve the problem from a reference frame attached to the railroad car. Relative to the railroad car, the ball has an initial velocity of 9.8 m/s in the vertically upward direction. Thereafter, the ball moves with acceleration g along the vertically downward direction and acceleration a along the horizontal direction. The time of flight of the ball is

$T=\frac{2(9.8\mathrm{m}/\mathrm{s})}{9.8\mathrm{m}/{\mathrm{s}}^2}=2 \mathrm{s}$

Horizontal displacement of the ball is given to be 2 m. Hence, we have 

$\Rightarrow \begin{array}{r}\frac{1}{2}a(2{)}^2=2\mathrm{m}\\ \Rightarrow a=1\mathrm{m}/{\mathrm{s}}^2\end{array}$