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A ball is thrown straight upward with a speed v from a point h meter above the ground. The time taken for the ball to strike the ground is

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$\frac{v}{g} [1 + \sqrt{1 + \frac{2 g h}{v^{2}}}]$

$\frac{v}{g} [1 - \sqrt{1 - \frac{2 g h}{v^{2}}}]$

$\frac{v}{g} [1 - \sqrt{1 + \frac{2 g h}{v^{2}}}]$

$\frac{v}{g} [2 + \frac{2 g h}{v^{2}}]$

answer is A.

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Detailed Solution

$- h = v t - \frac{1}{2} g t^{2} \Rightarrow \frac{1}{2} g t^{2} - v t - h = 0$

$t = \frac{v}{g} [1 + \sqrt{1 + \frac{2 g h}{v^{2}}}]$

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