Question

Shell is fired vertically upwards with a velocity ${v_1}$ ​ from a trolley moving horizontally with velocity ${v_2}$ a person on the ground observes the motion of the shell as a parabola, whose horizontal range is
A. $\dfrac{{2v_1^2{v_2}}}{g}$
B. $\dfrac{{2v_1^2}}{g}$
C. $\dfrac{{2v_2^2}}{g}$
D. $\dfrac{{2{v_1}{v_2}}}{g}$

Answer 1

Hint: Apply the kinematic equation of motion, where you substitute the value of $u$ is the initial velocity, $t$ is the time, $g$ is the acceleration of gravity. Then apply the range of the projectile formula. Then substitute the value of the velocities to determine the horizontal range of the projectile.

Formula used:
$R = \dfrac{{2{u_x}{u_y}}}{g}$
Where ${u_x},{u_y}$ are the initial velocity of vertical and horizontal components.

Complete step by step answer:
Kinematic equation of motion helps to describe a body’s location, velocity or acceleration relative to frame of reference. The velocity and the position can be derived from the newton equation by the method of integration. Here force acting on a body is known as the function of time. The velocity equation integration results in the distance equation

If the body is moving then resultant forces and moments can be non-zero. A force that makes a body move in a curved path is called centripetal force. Centripetal force is directed towards the centre and it acts on a body executing circular motion. In Free body diagram force is shown as a straight arrow pointing at the direction where force at.
Vertical velocity of the shell ${u_y} = {v_1}$
Shell’s horizontal velocity ${u_x} = {v_2}$
Range of the projectile $R = \dfrac{{2{u_x}{u_y}}}{g}$
Thus, we can write $R=\dfrac{{2{v_1}{v_2}}}{g}$.

Hence, the correct answer is option (D).

Note: The initial velocity will be zero if the motion starts from rest and the frame of reference should be the same. The velocity and the position can be derived from the newton equation by the method of integration. The velocity equation integration results in the distance equation.