A child bounces a 57 g superball on the sidewalk. The velocity change of the superball is from 24 m/s downward to 11 m/s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk? Answer in units of N.

Correct question:

A child bounces a 57 g superball on the sidewalk. The velocity change of the superball is from 24 m/s downward to 11 m/s upward. If the contact time with the sidewalk is 1/800 s, what is the magnitude of the average force exerted on the superball by the sidewalk? Answer in units of N.

The magnitude of the average force exerted on the superball by the sidewalk is 592.8 N

Explanation:

Given;

mass of the superball, m = 57 g = 0.057 kg

initial velocity of the superball, u = 24 m/s

final velocity of the superball, v = 11  m/s

contact time with the sidewalk, t =  1 / 800 s

To determine the magnitude of the average force exerted on the superball by the sidewalk, we apply Newton's second law of motion;

F = ma

Therefore, the magnitude of the average force exerted on the superball by the sidewalk is 592.8 N

lexy is right because she has the labeled s

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explanation: