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A stone is thrown vertically upward with a speed of $10 .0 {ms}^{- 1}$ from the edge of a cliff 65 m high what will be its speed just before hitting the bottom

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3.14 m/s

37.14 m/s

13.71 m/s

14.71 m/s

answer is B.

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Detailed Solution

H_{max} = \frac{u^{2}}{2g} = \frac{{(10)}^{2}}{2 (9.8)} = \frac{100}{19.6} = 5.1 m

$H_{max} from ground = 65 + 5.1$

$h = 70.1 m, u = 0$

$v = \sqrt{2 gh}$

$= \sqrt{2 (9.8) (7.1)}$

$v = 37.1 4 m/s$

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