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. u²/2g
• B. u/2g

Answer 1

The correct option is (B) $\frac{u^2}{2g}$

Given,

The initial velocity is $u$, Maximum height = h

Let the final velocity be $v$, acceleration be $a$ and distance covered be $s$.

We know from the equation of motion

Using the 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 the highest point, $v=0$

$\therefore$$0=u^2-2gh$

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

Hence the maximum height is $\frac{u^2}{2g}$