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A hiker stands on the edge of a cliff 495m above ground and throws a stone horizontally with a speed ‘u’. If it strikes the ground with a speed $100 m s^{- 1}$ then ‘u’ is
$[g = 10 {ms}^{- 2}]$

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$10 {ms}^{- 1}$

$4.95 {ms}^{- 1}$

$50 {ms}^{- 1}$

$20 {ms}^{- 1}$

answer is A.

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Detailed Solution

$v = \sqrt{u^{2} + 2 g h}$

$u^{2} = 100^{2} - 2 \times 495 \times 10 \Rightarrow u = 10 {ms}^{- 1}$

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