physics
posted by ShaSha .
A 3kg basketball is dropped from the top of a 120m building. if all of the energy is converted into KE, what is the velocity of the ball just before it hits the ground

potential energy = m g h = 3(9.81)(120)
Ke = .5 m v^2 = .5 (3)(v^2)
so
v^2 = 2 * 9.81 * 120 
I am lost...please explain

well, yo9u can just memorize the formula. Most of us know it.
For something dropped with no air resistance from height h
v = sqrt (2 g h) 
Let me see if I get this:
KE=1/2(m)(v^2)
but because I do not know the velocity, I would use the formula v=sqrt (2 g h)
v=sqrt (2 g h)
=sqrt (2*9.8*120)
=sqrt (2352)
then I would use the velocity 2352 for the formula
KE=1/2(m)(v^2)
=1/2 (3)(2352)
KE= 3528 
You could do it the way I did it:
Potential energy lost = kinetic energy gained
m g h = (1/2) m v^2
g h = (1/2) v^2
2 g h = v^2
v = sqrt (2 g h)
However many students and teachers have just plain memorized
v = sqrt (2 g h)
because it is used so frequently. 
Thank you so very much
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