If a body is thrown up with an initial velocity, $u$ and covers a maximum height, $h$ , 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,
Final velocity, $v = 0$

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

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

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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 $\frac{u^{2}}{2 g}$ . [ $g$ is the acceleration due to gravity]

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

If a body is thrown up with an initial velocity $u$ and covers a maximum height of $h$ , then $h$ is equal to

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If a body is thrown up with an initial velocity, u and covers a maximum height, h, then h is equal to u22g.

The correct option is A u2/2g Using equation, v2 = u2+2as Here, a=−g [∵ acceleration due to gravity is acting opposite to the direction of motion] s=h At highest point, Final velocity, v=0 ⇒0=u2−2gh ⇒h=u22g

If a body is thrown up with an initial velocity, $u$ and covers a maximum height, $h$ , 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,
Final velocity, $v = 0$

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

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