Question

kyle, a 100.0 kg football player, leaps straight up into the air (with no horizontal velocity) to catch a pass. he catches the 0.430 kg ball precisely at the peak of his jump, when he is 0.386 m off the ground. he hits the ground 0.0371 m away from where he leapt.
if the ball was moving horizontally when it was caught, how fast v was the ball traveling?

Explanation:

Step1: Calculate the time of fall

Use the equation \(h = \frac{1}{2}gt^{2}\) (where \(h = 0.386m\) and \(g=9.8m/s^{2}\)).

$$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times0.386}{9.8}}$$

$$t=\sqrt{\frac{0.772}{9.8}}\approx\sqrt{0.0788}\approx0.281s$$

Step2: Calculate the horizontal velocity of the combined system

The horizontal distance \(x = 0.0371m\). Using \(v_{combined}=\frac{x}{t}\)

$$v_{combined}=\frac{0.0371}{0.281}\approx0.132m/s$$

Step3: Use conservation of momentum

Let the mass of the player \(M = 100.0kg\), mass of the ball \(m = 0.430kg\). Initial momentum of the system (before catch) in horizontal direction: \(p_{initial}=mv\) (player has no horizontal velocity initially). After catch, \(p_{final}=(M + m)v_{combined}\)
By conservation of momentum \(mv=(M + m)v_{combined}\)

$$v=\frac{(M + m)v_{combined}}{m}$$

Substitute \(M = 100.0kg\), \(m = 0.430kg\), \(v_{combined}=0.132m/s\)

$$v=\frac{(100.0 + 0.430)\times0.132}{0.430}$$

$$v=\frac{100.43\times0.132}{0.430}=\frac{13.25676}{0.430}\approx30.8m/s$$

Answer:

\(30.8\)