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Q.

A stone is dropped from the top of a building. When it crosses a point 5 m below the top, another stone starts to fall from a point 25m below the top. Both stones reach the bottom of building simultaneously. The height of the building is

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a

45 m

b

50 m

c

25 m

d

35 m

answer is A.

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Detailed Solution

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$\begin{array}{l} time of release of 2^{nd} stone t = \sqrt{\frac{2 \times 5}{10}} = 1 \\ ∴ \frac{1}{2} g t^{2} = \frac{1}{2} g (t - 1)^{2} + 25 \\ \Rightarrow 5 t^{2} = 5 (t^{2} + 1 - 2 t) + 25 \\ 10 t = 30, t = 3 s, \\ h = \frac{1}{2} g t^{2} = \frac{1}{2} \times 10 \times 9 = 45 m \end{array}$

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