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:
[tex]p_{o} = p_{f} (1)[/tex]
- [tex]p_{o} = m_{K} *v_{oK} + m_{b} * v_{b} (2)[/tex]
- 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:
[tex]p_{f} = (m_{K} + m_{b}) * v_{fKb} (3)[/tex]
- 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:
[tex]v_{Kbf} =\frac{\Delta x}{\Delta t} (4)[/tex]
- 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:
- [tex]\Delta h = \frac{1}{2} * g *t^{2} = 0.386 m (5)[/tex]
[tex]t =\sqrt{\frac{\Delta h*2}{g}} = \sqrt{\frac{0.386m *2}{9.8m/s2} } = 0.28 s (6)[/tex]
- Replacing (6) and Δx in (4) we get the speed after the collision, as follows:
[tex]v_{Kbf} =\frac{\Delta x}{\Delta t} =\frac{0.0333m}{0.28s} = 0.12 m/s (7)[/tex]
- Replacing in (3), we have:
[tex]p_{f} = (m_{K} + m_{b}) * v_{fKb} = (85.0 kg + 0.430 kg) * 0.12 m/s = 10.3 kg*m/s[/tex]
- From (2) and (7), we can solve for vbo, as follows:
[tex]v_{bo} = \frac{p_{f} }{m_{b} } =\frac{10.3 kg*m/s}{0.430kg} = 24.0 m/s[/tex]
