Two stones are thrown up simultaneously from the edge of a cliff 240… | JEE Main 2015

Q46JEE Main 2015Difficult

Motion Under Gravity

Two stones are thrown up simultaneously from the edge of a cliff $240\text{ m}$ high with initial speed of $10\text{ m/s}$ and $40\text{ m/s}$ 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, take $g = 10\text{ m/s}^2$ )

(The figures are schematic and not drawn to scale)

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Answer: (a)

Both have the same acceleration, so $y_2 - y_1 = (40 - 10)t = 30t$ (straight line) while both are in flight.
Stone 1 lands: $-240 = 10t - 5t^2 \Rightarrow t = 8\text{ s}$ ; then $y_2 - y_1 = 240\text{ m}$ .
After that $y_1 = -240$ is fixed: $y_2 - y_1 = 40t - 5t^2 + 240$ , a downward-curving parabola.
Stone 2 lands when $t^2 - 8t - 48 = 0 \Rightarrow t = 12\text{ s}$ , where $y_2 - y_1 = 0$ .
So the answer is (a).

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