Question

A ball is dropped from the top of a building of height 300 M find velocity of the ball when potential energy remains 40% of its original value

Answer 1

Potential Energy (U) = mgh
when U becomes 40% of its original value so does its height
new height h'= (40/100)*300 = 120 m

Now by conservation of energy
Ui + Ki = Uf + Kf
mgh = mgh' + mv²/2
g(300) = g(120) +v²/2
v²= 3600
v = 60 m/s