Answer
See explanations.
Work Step by Step
Based on the given height equation, $s(t)=400-16t^2$, we know $s(0)=400\ ft$ and $s(t)=0$ gives the time when the camera hits the water; that is, $t=\sqrt {400/16}=5\ sec$. Use the limit of the difference quotient (or take the derivative of s(t) directly). We have the velocity $v(t)=s'(t)=-32t$ and at $t=5$, we have $v(5)=-32(5)=-160\ ft/sec$ when the camera hits the water, where the negative sign indicates that the velocity is downward.
