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(D) The shell is fired vertically upwards with velocity $v_1$ from a trolley moving horizontally with velocity $v_2$. Therefore,the initial velocity o

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$\frac{2 v_1 v_2}{g}$

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(D) The shell is fired vertically upwards with velocity $v_1$ from a trolley moving horizontally with velocity $v_2$. Therefore,the initial velocity of the shell relative to the ground has a horizontal component $u_x = v_2$ and a vertical component $u_y = v_1$.In the horizontal direction,there is no acceleration $(a_x = 0)$.In the vertical direction,the acceleration is due to gravity $(a_y = -g)$.The time of flight $T$ is determined by the vertical motion. The shell returns to the level of the trolley when the vertical displacement is zero:$y = u_y T - \frac{1}{2} g T^2 = 0$$v_1 T - \frac{1}{2} g T^2 = 0$$T = \frac{2 v_1}{g}$The horizontal range $R$ is the horizontal distance covered during the time of flight $T$:$R = u_x 	imes T$$R = v_2 	imes \frac{2 v_1}{g} = \frac{2 v_1 v_2}{g}$Thus,the correct option is $(d)$.

$A$ 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 ....

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