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Two stones are thrown simultaneously from the top of a cliff 100 m high, one upward with initial velocity 30 m/s and another vertically downward with an initial velocity 10 m/s. If air resistance is neglected, the separation between the stones 2 s later is (g=10 m/s²)

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$60 \sqrt{3} m$

zero

80 m

40 m

answer is D.

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Detailed Solution

$S_{1} = u_{1} t - \frac{1}{2} g t^{2}$

$= 30 \times 2 - \frac{1}{2} \times 10 \times 2^{2}$

= 40 m, upwards

$\begin{array}{l} S_{2} = u_{2} t + \frac{1}{2} g t^{2} \\ = 10 \times 2 + \frac{1}{2} \times 10 \times 2^{2} \\ = 40 downwards \\ \Rightarrow seperation=80 m \end{array}$

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