What is the expression for the maximum height reached by a body thrown vertically upwards with a velocity $u$ and time of flight $T$ .

h = \frac{g T^{2}}{8}

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h = \frac{u^{2}}{2 g}

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Both A & B

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None of the above

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Solution

The correct option is C Both A & B
When a body attains maximum height its final velocity $v = 0$ .
$v^{2} - u^{2} = - 2 g s$
$h = u^{2} / 2 g$

At highest point, $v = 0, t = T / 2$
$v = u - g t$
$T / 2 = u / g . . . . . . . . . . . (1)$
$s = u t - \frac{1}{2} g t^{2}$
$h = u T / 2 - \frac{1}{2} g (T / 2)^{2} . . . . . . . (2)$
From (1) and (2),
$h = g T^{2} / 8$

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If a body is thrown up with an initial velocity $u$ and covers a maximum height of $h$ , then $h$ is equal to $\frac{u^{2}}{2 g}$ :

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