1.

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»

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



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