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

The correct option is 4h.

To find the distance the body needs to fall for its velocity to double, we can use the equations of motion for a freely falling body.

Let the initial velocity of the body be $u=0$ (since it starts from rest), and the acceleration due to gravity be $g$.

According to the third equation of motion:
$v^2=u^2+2as$
where:
• $v$ is the final velocity,
• $u$ is the initial velocity,
• $a$ is the acceleration, and
• $s$ is the distance traveled.

Case 1: The body falls through a height $h$ to acquire a velocity $v$.
Substituting $u=0$, $a=g$, and $s=h$ into the equation of motion:
$v^2=0^2+2gh$
$v^2=2gh$ — (Equation 1)

Case 2: Let the total distance the body has to fall from rest for its velocity to become double ($2v$) be $H$.
Substituting the new final velocity $2v$ and distance $H$ into the equation of motion:
${\left(2v\right)}^2=0^2+2gH$
$4v^2=2gH$ — (Equation 2)

Now, substitute the value of $v^2$ from Equation 1 into Equation 2:
$4\left(2gh\right)=2gH$
$8gh=2gH$

Solving for $H$:
$H=\frac{8gh}{2g}$
$H=4h$

Therefore, the total distance the body has to fall from rest for its velocity to become double is $4h$.