Question

a body of mass M is dropped from a height h above the ground the velocity of the body when it has lost of its potential energy is​

Answer 1

lost of potential energy will be equal to gain in kinetic energy

when it drops from a hight of h

its velocity at ground is =(2gh)^1/2

gain in kinetic energy = 1/2Mv^2

= Mgh

Answer 2

Answer:

$V=\sqrt{2gh}\ m/s$

Explanation:

Given that

Mass = M

Height = h

A the initial position :

Potential energy = M g h

Kinetic energy = 0

At the final position :

Potential energy = 0

Kinetic energy = 1/2 M V²

So from energy consecration

M g h =  1/2 M V²

2 g h =  V²

$V=\sqrt{2gh}\ m/s$