Two stones are thrown up simultaneously from the edge of a cliff 240 m high with an initial speed of 10 m s$$^{-1}$$ and 40 m s$$^{-1}$$ respectively. Which of the following graph best represents the time variation of the relative position of the second stone with respect to the first? (Assume stones do not rebound after hitting the ground and neglect air resistance, take $$g = 10$$ ms$$^{-2}$$)(the figures are schematic and not drawn to scale)

1. When both stones are in air:

Position of stone 1: $$y_1 = u_1 t - \frac{1}{2}gt^2 = 10t - 5t^2$$

Position of stone 2: $$y_2 = u_2 t - \frac{1}{2}gt^2 = 40t - 5t^2$$

Relative position, $$(y_2 - y_1) = (40t - 5t^2) - (10t - 5t^2) = 30t$$

This is a straight line passing through the origin with a positive slope of $$30$$. This phase ends when the first stone hits the ground.

Time of impact:

For Stone 1: $$-240 = 10t - 5t^2 \implies t^2 - 2t - 48 = 0 \implies (t-8)(t+6)=0$$. Thus, $$t_1 = 8\text{ seconds}$$.

For Stone 2: $$-240 = 40t - 5t^2 \implies t^2 - 8t - 48 = 0 \implies (t-12)(t+4)=0$$.

Thus, $$t_2 = 12\text{ seconds}$$.

2. Stone 1 is on the ground, Stone 2 is in the air ($$8 < t \le 12\text{ s}$$)

Stone 1 is stationary: $$y_1 = -240\text{ m}$$.

Stone 2 continues moving: $$y_2 = 40t - 5t^2$$.

Relative position, $$(y_2 - y_1) = (40t - 5t^2) - (-240) = -5t^2 + 40t + 240$$

This is a downward-opening parabola.

At $$t = 8\text{ s}$$, $$y_2 - y_1 = 240\text{ m}$$.

At $$t = 12\text{ s}$$, $$y_2 - y_1 = 0\text{ m}$$.

Graph D correctly depicts the linear increase followed by a downward parabolic decrease.