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Chapter 4: Problem 15

After falling from rest from a height of \(30 \mathrm{~m}\), a \(0.50-\mathrm{kg}\) ball rebounds upward, reaching a height of \(20 \mathrm{~m}\). If the contact between ball and ground lasted \(2.0 \mathrm{~ms}\), what average force was exerted on the ball?

Short Answer

Expert verified
-11062.5 N

Step by step solution

01

Identify initial velocity after falling

The initial velocity of the ball right after falling can be obtained using the equation of motion \(v^2 = u^2 + 2as\), where \(u\) is the initial velocity (0), \(a\) is the acceleration due to gravity (9.8 m/s虏), and \(s\) is the distance fallen (30 m). Thus the velocity after falling is \(\sqrt{2 * 9.8 * 30} \approx 24.25 \, m/s\).
02

Identify final velocity after rebound

The final velocity before the ball starts to rise again can be obtained with the same equation of motion as before where \(s\) is now the height it reached after rebounding (20 m). We have \(v^2 = u^2 + 2as\), hence the velocity after rebound is \(-\sqrt{2 * 9.8 * 20} \approx -20 \, m/s\). The negative sign indicates that the direction of the velocity is upward.
03

Calculate change in velocity

The change in velocity is identified by subtracting the final velocity from the initial velocity. This gives \(-20 \, m/s - 24.25 \, m/s = -44.25 \, m/s\). The negative sign indicates that the velocity is in the opposite direction of the initial velocity.
04

Calculate average force

Finally, we use the second law of motion, \(F = ma\), where \(m\) is the mass (0.50 kg) and \(a\) is the acceleration. Here, acceleration is essentially the change in velocity over the change in time. The time of contact given is 2.0 ms. So, the average force exerted on the ball is \(F = m * \frac{\Delta v}{\Delta t} = 0.50 * \frac{-44.25}{2.0 * 10^{-3}} = -11062.5 \, N\). The negative sign indicates that the force is directed upward.
05

Conclusion

The average force exerted on the ball is 11062.5 N. This force is in the opposite direction to the force of gravity, hence it is exerted upward.

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