Posted by John on Thursday, May 12, 2011 at 10:02pm.
(1/2) m v^2 = m g h
v^2 = 2gh
v = sqrt(2gh) = 44.3 m/s
Your drag coef makes no sense to me. Usually
Drag force = (1/2) rho v^2 Cd A
where rho = density of fluid (air here)
Cd = drag coef
A = cross section area of ball (pi r^2)
so at terminal speed
m g = (1/2) rho Vt^2 Cd A
Calculate Vt
calculate (1/2) m Vt^2 = Ke
now the Ke at the bottom would be
m g h = m g(1300) if there were no friction
so
work done by friction = 1300 m g - (1/2) m Vt^2
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