Two stones are thrown up simultaneously from the edge of a cliff 240 m high with an initial speed of 10 m/s and 40 m/s respectively. Which of the following graph 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).

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Solution

Height of the cliff, h = 240 m
Velocity of Stone 1, $y_{1}$ = 10 m/s
Velocity of Stone 2, $y_{2}$ = 40 m/s
To calculate the time when stone 1 strikes the ground, use Equation of Motion,
$h = u t + \frac{1}{2} g t^{2}$
$- 240 = 10 \times t - \frac{1}{2} \times 10 \times t^{2}$
$5 t^{2} - 10 t - 240 = 0$
$t^{2} - 2 t - 48 = 0$
t = 8 sec and -6 sec
To calculate the time when stone 2 strikes the ground, use Equation of Motion,
$h = u t + \frac{1}{2} g t^{2}$
$- 240 = 40 \times t - \frac{1}{2} \times 10 \times t^{2}$
$5 t^{2} - 40 t - 240 = 0$
$t = 12 s e c$
The relative velocity of the two stone = $y_{2} - y_{1} = 40 - 10 = 30$ m/s
Therefore, graph of $y_{2} - y_{1}$ against time is a straight line.
Hence, when the stone strikes the ground, the relative velocity will increase up to 12 seconds.

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