Question

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

• A. True
• B. False

Answer 1

The correct option is A

True

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