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January 31, 2015

January 31, 2015

Posted by **John** on Thursday, May 12, 2011 at 10:02pm.

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 = ăg/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?

- Physics -
**Damon**, Thursday, May 12, 2011 at 10:18pm(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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