Question

if a body of mass 3 kg is dropped from top of a tower of height 50 m., then K.E.bafter 3 sec will be (ans=735J)

Answer 1

ANSWER:-
Here kinetic energy of a body is converted to half of its potential energy.
Therefore
K.E =1/2*m*g*h
Given mass m=3Kg
height h=50m
Therefore kinetic energy after 3 sec is 1/2 mgh
=1/2*3*9.8*50
=735 J
Hope it makes you clear.
Thank you.