Answer
2.98 m/s
Work Step by Step
As in Problem 14b, we use:
$v = \sqrt {v_{0}^{2} – 2g\Delta h}$
Here the change in height will be $\Delta h$ = 0, because the ball is at exactly the same height it was at 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:
$v = \sqrt {v_{0}^{2} – 2g\Delta h}$
= $\sqrt {(2.98 m/s)^{2} – 2(9.8 m/s²)(0)}$
= 2.98 m/s
