A body of mass m is released from the top of tower of height h. calculate its total energy at the top of the tower and then just before striking the ground . show that it is in accordance with the law of conservation of mechanical energy
Answers
Answered by
2
At the top of the tower
It's energy is= mgh
Just before striking the ground
It's energy is= 1/2mv^2.
It's energy is= mgh
Just before striking the ground
It's energy is= 1/2mv^2.
Similar questions