Question

A body freely falling from the rest has a velocity v after his falls through a height h. The distance, it has to fall down further for its velocity to become double, is :

• A. 4 h
• B. 2 s
• C. $1/\sqrt{2}s$
• D. $\sqrt{2}s$

Answer 1

The correct option is A 4 h

Here, initial velocity of the body $v_1=v$

Initial height h = hc

Final velocity of the body $v_2=2v$

Now, from the equation of motion

$v^2=u^2+2gh$

$v^2\propto h$

So, ${\left[\begin{array}{c}\frac{v_1}{v_2}\end{array}\right]}^2=\left[\begin{array}{c}\frac{h_1}{h_2}\end{array}\right]$

or, ${\left[\begin{array}{c}\frac{v}{2v}\end{array}\right]}^2=\frac{h}{h_2}$

or, $\frac{h}{h_2}=\frac{1}{4}$

or $h_2=4h$