A body, projected vertically upwards, crosses a point twice its journey at a height $h$ just after $t_{1}$ and $t_{2}$ seconds. Maximum height reached by the body is

\frac{g}{4} (t_{1} + t_{2})^{2}

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g {[\frac{t_{1} + t_{2}}{4}]}^{2}

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2 g {[\frac{t_{1} + t_{2}}{4}]}^{2}

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\frac{g}{4} (t_{1} t_{2})

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Solution

The correct option is D $2 g {[\frac{t_{1} + t_{2}}{4}]}^{2}$
$s_{A} = u t - \frac{1}{2} g t^{2} (a t t = t_{1}, t_{2})$
So, $s_{A} = u t_{1} - \frac{1}{2} g t_{1}^{2} - (1)$
$s_{A} = u t_{2} - \frac{1}{2} g t_{2}^{2} - (2)$
Equating $(1) a n d (2)$
$\Rightarrow u (t_{1} - t_{2}) = \frac{1}{2} g (t_{1}^{2} - t_{2}^{2})$
$\Rightarrow u = g \frac{(t_{1} + t_{2})}{2}$
$M a x h t = \frac{u^{2}}{2 g} = \frac{g^{2} (t_{1} + t_{2})^{2}}{4 \times 2 g} = 2 g (\frac{t_{1} + t_{2}}{4})^{2}$

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