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

u²/2g

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u/2g

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Solution

The correct option is A
u²/2g

Using equation $v^{2}$ = $u^{2} + 2 a s$
Here $a = - g$
[ $∵$ acceleration due to gravity is acting opposite to the direction of motion]
$s = h$

At highest point, $v = 0$

$0 = u^{2} - 2 g h$

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

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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 u22g.

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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 $\frac{u^{2}}{2 g}$ .

The correct option is A
u²/2g

Using equation $v^{2}$ = $u^{2} + 2 a s$
Here $a = - g$
[ $∵$ acceleration due to gravity is acting opposite to the direction of motion]
$s = h$

At highest point, $v = 0$

$0 = u^{2} - 2 g h$

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