Question

A body freely falling from the rest has a velocity $v$ after it falls through a height $h$. The distance, covered by body when it falls down with double of velocity, is :

• A. $12h$
• B. $8h$
• C. $9h$
• D. $4h$

Answer 1

Answer: (D)

From equation of motion,
we have $v^2=u^2+2gh$
where $u$ is initial velocity,
$v$ is final velocity and $h$ is height of fall.
Given, $u=0$ (body starts from rest).
$\therefore$ $v^2=2gh$
Given, $v_1=v,v_2=2v,h_1=h$
$\therefore \frac{{v_1}^2}{{v_2}^2}=\frac{2g{h}_1}{2g{h}_2}$
$\Rightarrow \frac{v^2}{4v^2}=\frac{h}{h_2}$
$\Rightarrow h_2=4h$ .