A boy standing in a trolley car projects a ball vertically upwards with respect to the trolley with a velocity ${v_1}$. If the trolley car moves with a constant velocity ${v_2}$ on a horizontal road then the correct option is
(A) range of the ball with respect to ground is $\dfrac{{2{v_1}{v_2}}}{g}$.
(B) time of flight of ball is $\dfrac{{2{v_1}}}{g}$
(C) maximum height of ball is $\dfrac{{2v_2^2}}{g}$
(D) both A and B

Hint: The range means that the distance covered by the ball when it is thrown from the moving trolley. The distance is the product of the velocity and the time. The time is determined by the velocity of the ball divided by the acceleration due to gravity on the ball.

Useful formula
The range of the ball with respect to the ground is,
$d = v \times t$
Where, $d$ is the range of the ball with respect to the ground, $v$ is the velocity of the car and $t$ is the time taken for the motion of the ball.
The time taken by the ball is,
$t = \dfrac{v}{g}$
Where, $t$ is the time taken for the motion of the ball, $v$ is the velocity of the ball thrown and $g$ is the acceleration due to gravity.

Complete step by step solution
Given that,
The velocity of the ball with respect to the trolley is ${v_1}$,
The velocity of the trolley car is ${v_2}$.
Now, the time of the motion can be formed from the object moving if the height is greater than zero,
Assume the height is zero, then the equation of height is written as,
$h = {v_1}t - \dfrac{{g{t^2}}}{2} = 0$
The above equation is also written as,
${v_1}t - \dfrac{{g{t^2}}}{2} = 0$
By rearranging the terms, then the above equation is written as,
${v_1}t = \dfrac{{g{t^2}}}{2}$
By cancelling the same terms, then the above equation is written as,
${v_1} = \dfrac{{gt}}{2}$
By rearranging the terms, then the above equation is written as,
$t = \dfrac{{2{v_1}}}{g}$
The range of the ball with respect to the ground is,
$d = v \times t$
Here the velocity is the velocity of the trolley car, so
$d = {v_2} \times t$
By substituting the time equation in the above equation, then
$d = {v_2} \times \dfrac{{2{v_1}}}{g}$
On multiplying the terms, then the above equation is written as,
$d = \dfrac{{2{v_1}{v_2}}}{g}$
Thus, the range of the ball with respect to the ground is $\dfrac{{2{v_1}{v_2}}}{g}$
The time of the flight of the ball is $\dfrac{{2{v_1}}}{g}$

Hence, the option (D) is the correct answer.

NoteThus, the ranger of the ball depends on both the velocity of the ball thrown in the vertical and the velocity of the trolley car moves in the horizontal and the acceleration of the ball due to gravity. Then the time taken of the ball depends on the velocity of the ball and the acceleration due to gravity on the ball.