Question

A body falling from rest has a velocity v after it falls through a height h. The distance it has to fall for its velocity to be 2v is.

Answer 1

The body is falling freely under gravity, so the gravitational acceleration = a = g

The initial velocity of the body = u = 0

After covering distance = h

The velocity becomes = v

So, v^2- u^2 = 2gh

or v^2 = 2gh

After sometime (t) its velocity becomes double = v' = 2v

Let the distance covered in time (t) = h'

Here initial velocity = v

So, v'^2 - v^2 = 2gh'

(2v)^2- v^2 = 2gh'

4v^2-v^2 = 2gh'

3v^2 = 2gh'

3(2gh) = 2gh'

3h = h'

Thus h' is 3 times h.
The distance the body has to fall for its velocity to become 2v from v is 3h

So the distance body has to fall for its velocity to become 2v from rest =
3h+h=4h