Posted by ShaSha on Monday, January 6, 2014 at 5:09pm.
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

physics  Damon, Monday, January 6, 2014 at 5:13pm
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

physics  ShaSha, Monday, January 6, 2014 at 5:34pm
I am lost...please explain

physics  Damon, Monday, January 6, 2014 at 5:40pm
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)

physics  ShaSha, Monday, January 6, 2014 at 5:50pm
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

physics  Damon, Monday, January 6, 2014 at 6:12pm
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.

physics  ShaSha, Monday, January 6, 2014 at 6:23pm
Thank you so very much
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