Asked by EMMANUEL
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 ?
Answers
Answered by
Henry
KE = 0.5M*V^2.
V^2 = Vo^2 + 2g*h = 0 + 19.6*3 = 58.8.
V^2 = Vo^2 + 2g*h = 0 + 19.6*3 = 58.8.
Answered by
bobpursley
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
Answered by
Henry
Yes, I agree.
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