Question

During basketball practice Shane made a jump shot, releasing a 0.60 kg basketball from his hands at a height of 2.0 m above the floor with a speed of 7.6 m/s. The ball swooshes through the net at a height of 3.0 m above the floor and with a speed of 4.5 m/s. How much energy was dissipated by air drag from the time the ball left Shane's hands until it went through the net?

Answer 1

5.373 J

Step-by-step explanation:

mass of ball (m) = 0.6 kg

initial height (h1) = 2 m

initial speed (u) = 7.6 m/s

final height (h2) = 3 m

final speed (v) = 4.5 m/s

acceleration due to gravity (g) = 9.8 m/s^{2}

from the conservation of energy ,

change in kinetic energy + change in potential energy = 0

in this case;

• change in kinetic energy = 0.5m
  `v^(2) -0.5mu^(2)`

= 0.5m(
`v^(2) -u^(2)`)

• change in potential energy = Uair + mg(h2-h1)
• Uair = energy dissipated by air drag

• the equation now becomes 0.5m(
  `v^(2) -u^(2)`)+Uair + mg(h2-h1)=0

Uair = -0.5m(
`v^(2) -u^(2)`) - mg(h2-h1)

Uair = -0.5x0.6(
`4.5^(2) -7.6^(2)`) - 0.6x9.8(3-2)

Uair = -(-11.253)- 0.6x9.8(3-2)

Uair = 5.373 J