A basketball player makers a jump shot. The 0.75 kg ball is released at a height

Question

A basketball player makers a jump shot. The 0.75 kg ball is released at a height of 1.50 m above the floor with speed of 6.80m/s                 

                    A. The ball goes through the net 2.60 m above the floor at a speed of 3.8 m/s. what is work done on ball by air resistance( nonconservative force)                 

                    B. calculate the maximum height that the ball could reach without air resistance                 

Explanation / Answer

Under ideal conditions (no air resistance involved), the velocity of the ball as it goes up to the net is (determined by the formula)

Vf^2 - Vi^2 = 2gs

where

Vf = ideal velocity of ball as it hits the net
Vi = initial velocity of release = 6.97 m/sec.
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
s = change in height = 3 - 2.11 = 0.89

Substituting values,

Vf^2 - 6.97^2 = 2(-9.8)(0.89)

NOTE the negative sign attached to the acceleration due to gravity. This merely implies that the ball is slowing down as it is going up.

Vf^2 =31.14

Vf = 5.58 m/sec.

Since the actual velocity of the ball is 4.21 meters, then the work done by air resistance is

Wr = difference in kinetic energy

Wr = (1/2)(0.6)[5.58^2 - 4.21^2)

Wr = 4.02 joules

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