The height from which the body is released is numerically equal to the velocity accquired find the that height is equal to ? Ans:2g

Oct 5, 2017

This is just algebra, but isn't simple...

Explanation:

Due to conservation of energy, ${E}_{p}$ lost = ${E}_{k}$ gained

So $m . g . h = \frac{1}{2} m {v}^{2}$

So $g . h = \frac{1}{2} {v}^{2}$

Or 2g.h = v^2

If we say v = h then substitute that into the previous equation,

$2 g . h = {h}^{2}$

So dividing both sides by h get us

2g = h

That means if we drop an object of any mass from 19.82 m it will hit the ground at 19.82 m/s.