Work done by air resistance on a basketball shot

[Bcisewski]

Formula assistance, can anyone assist?

A basketball player makes a jump shot. The 0.534-kg ball is released at a height of 2.01 m above the floor with a speed of 8.95 m/s. The ball goes through the net 3.48 m above the floor at a speed of 4.87 m/s. What is the work done on the ball by air resistance, a nonconservative force?

 

[Pyrrhus]

Help yourself with

Work-Kinetic energy theorem

$$\sum_{i=1}^{n} W_{i} = \Delta K$$

 

[wais]

To calculate the work done by air resistance, we can use the formula W = Fd, where W is the work done, F is the force of air resistance, and d is the distance traveled by the ball.

First, we need to find the force of air resistance. This can be calculated using the formula F = 0.5 * p * v^2 * A, where p is the density of air, v is the velocity of the ball, and A is the cross-sectional area of the ball.

Assuming a density of air of 1.2 kg/m^3 and a cross-sectional area of the ball of 0.03 m^2, we can calculate the force of air resistance as follows:

F = 0.5 * 1.2 kg/m^3 * (4.87 m/s)^2 * 0.03 m^2
= 0.353 N

Next, we need to find the distance traveled by the ball. This can be calculated by subtracting the initial height from the final height:

d = 3.48 m - 2.01 m
= 1.47 m

Now, we can plug in our values into the work formula:

W = Fd
= (0.353 N)(1.47 m)
= 0.52 J

Therefore, the work done by air resistance on the basketball is approximately 0.52 J. This means that the air resistance has slowed down the ball's motion and caused it to lose some of its energy.