(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
I've tried a bunch, have no idea. Got the first part, no idea on the second one.
A basketball, of mass m = 0.45 kg is dropped, from rest, from a stationary hellicopter at a high altitude. Use g = 9.81 N/kg.
(a) If air resistance is negligible, using energy considerations alone, how fast should the ball be going after it falls 100 meters? 1 m/s
(b) Of couse, we cannot ignore air resistance (i.e. drag). As a result, after the ball has fallen far enough, it reaches its terminal velocity, vt = �ãg/d, where d is the drag coefficient of the ball in air. If d = 0.007, how much energy has the ball lost due to air resistance after it falls 1300 meters?
1 answer
1 answer