Question

Two stones are thrown up simultaneously from the edge of a cliff $240\, m$ high with initial speed of $10 \,m/s$ and $40 \,m/s$ respectively. Which of the following graphs best represents the time variation of 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 $g = 10 \,m/s^2)$ (The figures are schematic and not drawn to scale.)

• A.
• B.
• C.
• D.

Answer 1

Answer: (C)

For the second stone time required to reach the ground
is given by $y = ut - \frac{1}{2} gt^2$
$-240 = 40t - \frac{1}{2} \times 10 \times t^2$
$\therefore 5t^2 - 40t -240 = 0$
$(t - 12)( t + 8 ) = 0$
$\therefore t = 12\,s$
For the first stone
$ - 240 = 10 t - \frac{1}{2} \times 10 \times t^2$
$\therefore - 240 = 10 t - 5t^2$
$ 5t^2 - 10t - 240 = 0$
$(t - 8) ( t + 6) = 0$
$T = 8\,s$
During first $8$ seconds both the stones are in air:
$\therefore y_2 - y_1 = (u_2 - u_1)t = 30 t$
So, graph of $(y_2 - y_1)$ against $t$ is a straight line.
After $8$ seconds,
$y_2 = u_2t - \frac{1}{2} gt^2 - 240$
Stones two has acceleration with respect to stone one.