Question

A basketball player makes a jump shot. The 0.630-kg ball is released at a height of 1.80 m above the floor with a speed of 7.09 m/s. The ball goes through the net 3.03 m above the floor at a speed of 4.21 m/s. What is the work done on the ball by air resistance, a nonconservative force

Answer 1

mass of the ball m = 0.63 kg

initial height h = 1.8 m

final height h ' = 3.03 m

initial speed v = 7.09 m / s

final speed v ' = 4.21 m / s

Let the work done on the ball by air resistance W = ?

we know from law of conservation of energy ,

total energy at height h + work done by air = total energy at height h '

  mgh + ( 1/ 2) mv^ 2 + W = mgh ' + ( 1/ 2) mv'^ 2

0.630*9.8*1.8 + 0.63*7.09^2 + W = mgh ' + ( 1/ 2) mv'^ 2

From there you can find W

if there is negative sign indicates it work opposite direction to motion