(1) The work done by Howard is 9 J.
(2) The change in kinetic energy of the ball is 4.9 J.
(3) After the ball was launched, the work done on the ball was the change in kinetic energy of the ball = 4.9 J.
(4) Gravity did the work
<h3>
Work done by Howard</h3>
The work done by Howard can be calculated using the principle of conservation of energy.
Work done by Howard = Kinetic energy of the basketball
K.E = ¹/₂mv²
K.E = ¹/₂(0.5 kg)(6 m/s)²
K.E = 9 J
<h3>Change in kinetic energy of the ball after landing</h3>
ΔK.E = K.E(final) - K.E(initial)
ΔK.E = ¹/₂mvf² - ¹/₂mvi²
ΔK.E = ¹/₂m(vf² - vi²)
ΔK.E = ¹/₂(0.5)(7.46² - 6²)
ΔK.E = 4.9 J
After the ball was launched, the work done on the ball was the change in kinetic energy of the ball = 4.9 J.
Gravity did the work, since change in kinetic energy = gravitational energy.
Learn more about kinetic energy here: brainly.com/question/25959744
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