Velocity at a given time

Differentiate $s(t) = -16t^2 + 64t + 5$ term by term using the power rule: the derivative of $-16t^2$ is $-32t$, the derivative of $64t$ is $64$, and the derivative of the constant $5$ is $0$. This gives $v(t) = s'(t) = -32t + 64$.

Substitute $t = 1$ into $v(t)$: $v(1) = -32(1) + 64 = 32$. Since this is positive, the projectile is moving upward at $t=1$ with a velocity of $32$ ft/s.

The maximum height occurs when the velocity is zero, since the projectile transitions from moving upward (positive velocity) to moving downward (negative velocity) at that instant. Using $v(t) = -32t + 64$, set $v(t) = 0$.

Solving $-32t + 64 = 0$ gives $t = 2$. Therefore, the projectile reaches its maximum height at $t = 2$ seconds.