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A particle is projected vertically upwards from a point A on the ground. It takes $t_{1}$ time to reach a point B but it continues to move up. If it takes further $t_{2}$ time to reach the ground from point B, then height of point B from the ground is

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$\frac{1}{8} g {(t_{1} + t_{2})}^{2}$

$\frac{1}{2} g t_{1} t_{2}$

$g t_{1} t_{2}$

$\frac{1}{2} g {(t_{1} + t_{2})}^{2}$

answer is D.

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Detailed Solution

$h = u t_{1} - \frac{1}{2} g t_{1}^{2}$

$\begin{array}{l} \frac{2 u}{g} = t_{1} + t_{2} \\ So h= \frac{g}{2} t_{1} t_{2} \end{array}$

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