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A stone is dropped from the top of a building of height h. After 1 s, another stone is dropped from the balcony 25 m below the top of the building. If both the stones strike the ground at the same time, what is value of h? Take g = 10 ms^-2

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45 m

75 m

30 m

60 m

answer is B.

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Detailed Solution

$h = \frac{1}{2} g t^{2} (1) h - 25 = \frac{1}{2} g {(t - 1)}^{2} (2)$ .

Subtracting (2) from (1), we get

$25 = \frac{1}{2} g (2 t - 1) = \frac{1}{2} \times 10 \times (2 t - 1)$

which gives t= 3 s. Using t= 3 s in (1), we get h = 45 m.

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