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A ball of mass I .5 kg falls vertically downward. Just before striking the floor, its speed is 14 m/s. Just after rebounding upward, its speed is 10 m/s.
If this change of velocity took place in 0.20 seconds, what is the average force of the ball on the floor?

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opudodennis

Answer:

180 N

Explanation:

We know that acceleration is the rate of change of speed per unit time hence

[tex]a=\frac {v_f-v_i}{t}[/tex] where v and t are velocity and time respectively, f and i represent final and initial.

Also, from Newton's law of motion, F=ma and replacing a with the above then

[tex]F=m\frac {v_f-v_i}{t}[/tex]

Substituting 1.5 Kg for mass, m -14 m/s for i and 10 m/s for for v then

[tex]F=1.5\times \frac {10--14}{0.2}=180 N[/tex]

Therefore, the force is 180 N

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