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A stone is allowed to fall from the top of a tower 300m height and at the same time another stone is projected vertically upward from the ground with a velocity $100 m s^{- 1}$ . The height above the ground where the two stones meet is

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44.1m

255m

44m

255.9m

answer is D.

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Detailed Solution

$\begin{array}{l} x = 100 t - \frac{1}{2} g t^{2} .... (1) \\ 300 - x = 0 + \frac{1}{2} g t^{2} ...... (2) \\ S o l v e, 1 & 2 \\ t = 3 \sec, x = 255.9 m \end{array}$

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