Answer
$-16 \sqrt {125}$
Work Step by Step
When the can hits the ground at the moment $t=a$ , then $s(a)=0 $
This implies that $ -16a^2+500=0$
After solving we get $a=\sqrt{\dfrac{125}{4}}$ ...(1)
Now, the velocity at that moment is: $v(a)=\lim\limits_{t \to a}\dfrac{s(t)-s(a)}{t-a}=\lim\limits_{t \to a}\dfrac{-16t^2+500+16a^2+500}{t-a}$
or, $\lim\limits_{t \to a}\dfrac{-16t^2+500+16a^2+500}{t-a}=-16\lim\limits_{t \to a}\dfrac{(t-a)(t+a)}{t-a}=-32a$
Fro equation (1), we have $v(a)=-32a$
From equation (1), we get
$v(a)=(-32) \sqrt{\dfrac{125}{4}}=-16 \sqrt {125}$
