Question

. 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.

Answer 1

so whats your question?

Answer 2

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.

Answer 3

I also don't know how to do it. To be honest, it is because I did not understand the question...