KE = 0.5M*V^2.
V^2 = Vo^2 + 2g*h = 0 + 19.6*3 = 58.8.
A mass of 2kg is dropped from a height of 3m.Neglecting air resistance,KE of the stone just b4 it hits the ground is ?
3 answers
and of course, the final KE is equal to the initial PE...
initial PE=mgh=2*9.8*3=58.8Joules
initial PE=mgh=2*9.8*3=58.8Joules
Yes, I agree.