Question

A body of mass m is dropped from q height h derive an expression for kinetic energy of the body when it reaches the ground

Answer 1

Answer:

Mgh

Explanation

formula of final velocity when a body is dropped from tower of height h is (2gh)*1/2

k

$kinetic \: energy \: = \frac{1}{2} m {v}^{2}$

=

$= \frac{1}{2} m { \sqrt{2gh} }^{2}$

= Mgh