A body is projected vertically upwards at time $t = 0$ and it is seen at a height $H$ at times $t_{1}$ and $t_{2}$ seconds during its flight. The maximum height attained is ( $g$ is the acceleration due to gravity)

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

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

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

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

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Solution

The correct option is C $\frac{g (t_{1} + t_{2})^{2}}{8}$
Time taken for the body to reach the highest point is $\frac{(t_{1} + t_{2})}{2}$
As $v = u - g t$ ;
At highest point, $v = 0$ ;
Therefore, initial velocity of the body is $u = \frac{g (t_{1} + t_{2})}{2}$
Maximum height attained by the body,
$H_{m a x} = \frac{u^{2}}{2 g} = \frac{1}{2 g} {(\frac{g (t_{1} + t_{2})}{2})}^{2}$
$H_{m a x} = \frac{g (t_{1} + t_{2})^{2}}{8}$

Alternative :
For a body projected vertically upwards,
$h = u t - \frac{1}{2} g t^{2}$
If $t_{1}$ and $t_{2}$ are the roots of the equation,
$t_{1} + t_{2} = \frac{2 u}{g}$ and $t_{1} t_{2} = \frac{2 h}{g}$
$\Rightarrow u = \frac{g (t_{1} + t_{2})}{2}$
Therefore maximum height attained,
$H_{m a x} = \frac{u^{2}}{2 g} = \frac{g (t_{1} + t_{2})^{2}}{8}$

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A body is projected vertically upwards at time t=0 and it is seen at a height H at times t1 and t2 seconds during its flight. The maximum height attained is (g is the acceleration due to gravity)

A body is projected vertically upwards at time $t = 0$ and it is seen at a height $H$ at times $t_{1}$ and $t_{2}$ seconds during its flight. The maximum height attained is ( $g$ is the acceleration due to gravity)