## Fundamentals of Physics Extended (10th Edition)

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