Answer
(a) $-64$ ft/s
(b) $-32a$ ft/s
(c) $6.32$ seconds
(d) $-202.4$ ft/s
Work Step by Step
(a) Given the equation for the height $h(t)=640-16t^2$, the velocity at $t$ is given by the derivative of the height with respect to $t$, we have:
$v(t)=h'(t)=\lim_{k\to 0}\frac{h(t+k)-h(t)}{k}=\lim_{k\to 0}\frac{640-16(t+k)^2-(640-16t^2)}{k}=\lim_{k\to 0}\frac{-16(t^2+2kt+k^2)+16t^2}{k}=\lim_{k\to 0}\frac{-32kt-16k^2}{k}=\lim_{k\to 0}(-32t-16k)=-32t$
Thus we have $v(2)=-32\times2=-64$ ft/s downward.
(b) At $t=a$, we have $v(a)=-32a$ ft/s downward.
(c) When the stone hits the ground $h(t)=0$, we have $640-16t^2=0$ which gives $t^2=40$ or $t=2\sqrt {10}\approx6.32$ seconds
(d) When the stone hits the ground, $t=2\sqrt {10}$ and $v(t)=-32\times2\sqrt {10}\approx-202.4$ ft/s downward.
