Average velocity on [1,3] via the secant line slope

First evaluate the height function at the endpoints of the interval. We have $s(1) = -16(1)^2 + 64(1) + 5 = -16 + 64 + 5 = 53$, and $s(3) = -16(3)^2 + 64(3) + 5 = -144 + 192 + 5 = 53$.

The average velocity is the average rate of change, given by $\dfrac{s(3) - s(1)}{3 - 1} = \dfrac{53 - 53}{2} = \dfrac{0}{2} = 0$. So the average velocity over $[1,3]$ is $0$ feet per second, meaning the projectile returns to the same height at $t=3$ as it had at $t=1$.