Question

When a body is projected vertically up from the ground its velocity is reduced to $\frac{1}{4}$th of its velocity at ground at height $h$. Then the maximum height reached by the body is

• A. $\frac{15}{32}h$
• B. $\frac{15}{16}h$
• C. $\frac{15}{8}h$
• D. $\frac{5}{4}h$

Answer 1

The correct option is B $\frac{15}{16}h$

Using the equation of motion ,
$v^2=u^2-2gh$
$\frac{u^2}{16}=u^2-2gh$
$h=\frac{15u^2}{32g}$
At Max. height, $v=0$ and equation of motion is
$v^2=u^2-2gh$
$H=\frac{u^2}{2g}=\frac{15h}{16}$