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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