Answer
4.21 m/s
Work Step by Step
As in Problem 14a, we have that $\frac{1}{2} m(v^2–v_{0}^{2})+mgΔh=0$.
Solving for $v$, we get:
$v = \sqrt {v_{0}^{2} – 2g\Delta h}$
Here the change in height will be $\Delta h$ = –0.452 m, because the ball drops to the lowest point, which is a vertical distance L = 0.452 m from where it was initially. The initial speed will be the one found in Problem 14a, $v_{0}$ = 2.98 m/s.
So plugging in our values, we get:
$v = \sqrt {v_{0}^{2} – 2g\Delta h}$
= $\sqrt {(2.98 m/s)^{2} – 2(9.8 m/s²)(–0.452 m)}$
$\approx$ 4.21 m/s
