From v^2 =u^2 +2ghU=0 so v= under root 2ghAfter falling h hight it's velocity is V=v2gh..... 1After velocity becomes double 2V=v2gh' ..2 small v is underootPut value of V from eqn 1 to eqn2Therefore answer will be h'=4h

Introduction:
When a body falls from rest, it accelerates due to the force of gravity. As it falls through a certain height, it gains velocity. The question asks for the distance the body needs to fall further in order for its velocity to become double.

Understanding the Problem:
To solve this problem, we need to apply the equation of motion for a falling body.

Equation of Motion:
The equation of motion for a falling body is given by:
v^2 = u^2 + 2gh

Where:
v = final velocity
u = initial velocity (which is zero in this case)
g = acceleration due to gravity
h = height fallen

Calculating the Distance:
We are given that the body falls through a height h and its velocity becomes V. We need to find the distance it needs to fall further for its velocity to become double.

Let's assume the distance the body needs to fall further is d.

Therefore, the final velocity after falling through the distance d will be 2V.

Using the equation of motion, we can write:
(2V)^2 = 0^2 + 2g(h + d)

Simplifying the equation, we get:
4V^2 = 2gh + 2gd

Now, let's solve for d:

4V^2 - 2gh = 2gd

Dividing both sides by 2g, we have:
2V^2 - h = gd

Finally, solving for d, we get:
d = (2V^2 - h) / g

Conclusion:
The distance the body needs to fall further for its velocity to become double is given by (2V^2 - h) / g. This equation represents the additional height the body needs to fall in order to gain double the velocity it initially had after falling through a height h.