Question

Kyle, a 85.0 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 meters off the ground. He hits the ground 0.0333 meters away from where he leapt. If the ball was moving horizontally when it was caught, how fast was the ball traveling

Answer 1

Answer:

v = 24.0 m/s

Explanation:

• When Kyle catches the pass, there is an inelastic collision between him and the ball, since both move together after it.
• Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

       $p_{o} = p_{f} (1)$

• $p_{o} = m_{K} *v_{oK} + m_{b} * v_{b} (2)$

• Since Kyle catches the ball at the peak of his jump, his speed at the time of the collision is just zero, so voK =0 in (2).
• The final momentum is just the product of the sum of both masses times the final speed of both objects moving together after the collision, as follows:

       $p_{f} = (m_{K} + m_{b}) * v_{fKb} (3)$

• Now, we have two unknowns yet: vbo (our target) and vfKb.

• Vfkb, is just the horizontal speed after the collision, which is constant, assuming no other forces acting on Kyle than gravity.
• We can apply then the definition of average speed, as follows:

        $v_{Kbf} =\frac{\Delta x}{\Delta t} (4)$

• where Δx = 0.033 m (given).
• The time after the collision can be obtained knowing that Kyle is in free fall since the collision till he hits the ground, starting from rest, at 0.386 m from ground.
• So, we can apply the kinematic equation for the vertical displacement, as follows:
• $\Delta h = \frac{1}{2} * g *t^{2} = 0.386 m (5)$

• Solving for t, we get:

        $t =\sqrt{\frac{\Delta h*2}{g}} = \sqrt{\frac{0.386m *2}{9.8m/s2} } = 0.28 s (6)$

• Replacing (6) and Δx in (4) we get the speed after the collision, as follows:

        $v_{Kbf} =\frac{\Delta x}{\Delta t} =\frac{0.0333m}{0.28s} = 0.12 m/s (7)$

• Replacing in (3), we have:

        $p_{f} = (m_{K} + m_{b}) * v_{fKb} = (85.0 kg + 0.430 kg) * 0.12 m/s = 10.3 kg*m/s$  

• From (2) and (7), we can solve for vbo, as follows:

       $v_{bo} = \frac{p_{f} }{m_{b} } =\frac{10.3 kg*m/s}{0.430kg} = 24.0 m/s$