Answer:
(a) The coefficient of restitution is 0.8.
(b) The ball must strike the surface with a 50.14 ft/s vertical velocity, approximately.
Explanation:
The ratio between the final and the initial height is given by
[tex]h_f = 0.64 h_i[/tex]
Before the impact, we have
[tex]v_f^2 = v_0^2 + 2gh_i\\v_f = \sqrt{2gh_i}.[/tex]
Analogously after the impact
[tex]v_f^2 = v_0^2 - 2gh_f\\v_0 = \sqrt{2gh_f}.[/tex]
From the previous, the restitution coefficient can be calculated from the ratio between the ball's velocity, after and before the impact, i.e.
[tex]\frac{v_0}{v_f} = \frac{\sqrt{2gh_f}}{\sqrt{2gh_i}} = \sqrt{\frac{h_f}{h_i}} = \sqrt{\frac{0.64 h_i}{h_i}} = \sqrt{0.64} = \mathbf{0.8}.[/tex]
If the ball is meant to rebound to a 25 ft height, we plug in this value in one of the previous equations as follows
[tex]v_f = \sqrt{2gh_i} = \sqrt{2g\frac{h_f}{0.64}} = \sqrt{2\times 32.18\frac{ft}{s^2} \times\frac{25 ft}{0.64}} \approx \mathbf{50.14 ft/s}.[/tex]
