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 1 m/s² and the projection velocity in the vertical direction is 9.8 m/s. How far behind the boy will the ball fall on the car?

Given:
Acceleration of the car = 1 m/s²
Projection velocity of the ball (considered as a projectile) in the vertical direction = 9.8 m/s
Angle of projection, α = 90˚
Let u be the initial velocity of the car when the ball is thrown.
Both the car and the ball have the same horizontal velocity.
We know that the distance travelled by the ball in horizontal direction is given by
s = ut
Here, t is the time.
Also, the distance travelled by the car in horizontal direction is given by
$s\text{'}=ut+\frac{1}{2}a{t}^2$
Time of flight of the projectile:
$t=\frac{2u\sin \alpha }{g}$
g = 9.8 m/s²
$\Rightarrow t=\frac{2\times 9.8}{9.8}=2\mathrm{s}$
Distance between the accelerated car and the projectile:
$s\text{'}-s=\frac{1}{2}a{t}^2=\frac{1}{2}\times 1\times 2^2=2\mathrm{m}$
Therefore, the ball drops 2 m behind the boy.