Question

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

• A. u²/2g
• B. u/2g

Answer 1

The correct option is A

u²/2g

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

At highest point $v=0$

$0=u^2-2gh$

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