This is a related rates problem with a twist.

Suppose you have a street light at a height $H$. You drop a rock vertically so that it hits the ground at a distance $d$ from the street light. Denote the height of the rock by $h$. The shadow of the rock moves along the ground. Let $s$ denote the distance of the shadow from the point where the rock impacts the ground. Of course, $s$ and $h$ are both functions of time. To enter your answer into WeBWorK use the notation $v$ to denote $h'$: $$v=h'.$$ Then the speed of the shadow at any time while the rock is in the air is given by $s'=$ (where $s'$ is an expression depending on $h$, $s$, $H$, and $v$ (You will find that $d$ drops out of your calculation.) Now consider the time at which the rock hits the ground. At that time $$h=s=0.$$ The speed of the shadow at that time is $s'=$ where your answer is an expression depending on $H$, $v$, and $d$.
Hint: Use similar triangles and implicit differentiation. For the second part of the problem you will need to compute a limit.