Standing on a long railroad car, a boy throws a ball straight...

The time of flight of the ball is given by $T = \dfrac{2u_y}{g}$.

Substituting $u_y = 9.8 \text{ m/s}$ and $g = 9.8 \text{ m/s}^2$, we get $T = \dfrac{2 \times 9.8}{9.8} = 2 \text{ s}$.

In the horizontal direction, the ball moves with a constant velocity equal to the velocity of the car at the instant of projection.

The car accelerates horizontally with $a = 1 \text{ m/s}^2$.

The relative horizontal acceleration of the ball with respect to the car is $a_{rel} = -1 \text{ m/s}^2$.

The initial relative horizontal velocity is $u_{rel} = 0$.

The horizontal displacement of the ball relative to the car is $s = u_{rel}T + \dfrac{1}{2}a_{rel}T^2$.

Substituting the values, $s = 0 + \dfrac{1}{2}(-1)(2)^2 = -2 \text{ m}$.

The ball will fall $2 \text{ m}$ behind the boy.