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const2013 [10]
3 years ago
14

A 0.140-kg baseball is dropped from rest. It has a speed of 1.20 m/s just before it hits the ground, and it rebounds with an upw

ard speed of 1.00 m/s. The ball is in contact with the ground for 0.0140 s.
Required:
What is the average force exerted by the ground on the ball during this time? Also explain whether it's upwards or downwards.
Physics
1 answer:
irga5000 [103]3 years ago
6 0

Answer:

22 N upward

Explanation:

From the question,

Applying newton's second law of motion

F = m(v-u)/t....................... Equation 1

Where F = Average force exerted by the ground on the ball, m = mass of the baseball, v = final velocity, u = initial velocity, t = time of contact

Note: Let upward be negative and downward be positive

Given: m = 0.14 kg, v = -1.00 m/s, u = 1.2 m/s, t = 0.014 s

Substitute into equation 1

F = 0.14(-1-1.2)/0.014

F = 0.14(-2.2)/0.014

F = 10(-2.2)

F = -22 N

Note the negative sign shows that the force act upward

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