After falling from rest at a height of 32.8 m, a 0.594 kg ball rebounds upward,

Question

After falling from rest at a height of 32.8 m, a 0.594 kg ball rebounds upward, reaching a height of 19.7 m. If the contact between ball and ground lasted 2.35 ms, what average force was exerted on the ball? (in N)
A: 1.22×103 B: 1.77×103 C: 2.57×103 D: 3.73×103 E: 5.41×103 F: 7.85×103 G: 1.14×104 H: 1.65×104
After falling from rest at a height of 32.8 m, a 0.594 kg ball rebounds upward, reaching a height of 19.7 m. If the contact between ball and ground lasted 2.35 ms, what average force was exerted on the ball? (in N)
A: 1.22×103 B: 1.77×103 C: 2.57×103 D: 3.73×103 E: 5.41×103 F: 7.85×103 G: 1.14×104 H: 1.65×104
After falling from rest at a height of 32.8 m, a 0.594 kg ball rebounds upward, reaching a height of 19.7 m. If the contact between ball and ground lasted 2.35 ms, what average force was exerted on the ball? (in N)
A: 1.22×103 B: 1.77×103 C: 2.57×103 D: 3.73×103 E: 5.41×103 F: 7.85×103 G: 1.14×104 H: 1.65×104

Explanation / Answer

Before impact, v = -(2gh) = -(2 * 9.8m/s² * 32.8m) = -25.35m/s
After impact, v = (2 * 9.8m/s² * 19.7m) = 19.65 m/s
v = 45 m/s
Favg = mv / t = 0.594kg * 45m/s / 0.00235s = 11374.5 N=1.14*10^4 N

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