(1/2) m v^2 = m g h
v = sqrt (2 g h)
= sqrt (2*9.81*15.5)
= 17.4 m/s
if v = 8
Ke = (1/2)(.14)(64) = 4.48 Joules
Pe at top = m g h = .14*9.81*15.5
= 21.3 Joules
energy lost = work done by friction = 21.3-4.48 = 16.8 Joules
Force * distance = 16.8
F = 16.8/15.5 = 1.08 Newtons
A 140 g baseball is dropped from a tree 15.5 m above the ground. With what speed would it hit the ground if air resistance could be ignored? If it actually hits the ground with a speed of 8.00 m/s, what is the magnitude of the average force of air resistance exerted on it?
2 answers
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2 answers