A light shines from the top of a pole 50 feet high. A ball is dropped from the same height from a .30 feet away from the light. How fast is the shadow of the ball moving along the ground one second later? Assume the ball falls a distance S = 16 T squared in T seconds. So if S = 16 T squared, and we're saying it's one second later, we know that it's fallen. 16 S equals 16 S equals 16 T squared. We're going to find the derivative of that distance in terms of time. So DSDT is going to equal 32TD, TDTDTDT is just one. The time is 1, so our DSDT is going to be 32 feet per second. Now we have similar triangles. If we look at the big triangle, we see it's 50 on the height and X on the length. Now this other triangle, this portion here is going to be the whole distance, the 50 minus how much it's dropped. So 50 -, s is going to be this piece in here. And then this piece, the base was the entire X minus the 30 that we're not going to be including in the new triangle. So by similar triangles, we know that the base to the side X to 50 is going to equal the base to the side X -, 30 / 50 -, s So if we cross multiply, we get S, we get X, we get 50X minus XS equaling fifty X -, 1500. If we solve for X, the 50 XS on both sides are going to cancel if we divide by a negative S negative divided by negative. So we get 1500 / s Now we're going to take the derivative. So DX DT is going to equal -1500 S squared DS DT. The S was 16 at one second, so we're going to put in 16 for the S. The DSDT we found is 32 feet per second. So when we compute this, we're going to get -375 / 2 feet per second. So the shadow of the ball moving along the ground one second later, how fast it's moving is -37375 / 2 feet per second.