Question

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

• A. $\frac{u}{2g}$.
• B. $\frac{u^2}{2g}$.
• C. 0
• D. $\frac{u^2}{g}$.

Answer 1

The correct option is B

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

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 $v=0$

$0=u^2-2gh$

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