A light is at the top of a pole 80ft high. A ball is dropped at the same height from a point 20ft from the light. If the ball falls according to 2=16t^2, how fast is the shadow of the ball moving along the ground 1 second later?

Question

please help!!!

Let x be the distance of the shadow of the ball from the pole, on the ground. Draw yourself a picture and consider the two similar right triangles with side lengths (y,20) and (80,x).

y = the distance fallen, 16 t^2 (feet)

Clearly,
x/80 = 20/y
x = 1600/16 t^2 = 100/t^2
Solve for dx/dt @ t=1 s

Answer 1

dx/dt = -200/t^3
At t=1 s, dx/dt = -200/1^3 = -200 ft/s

Therefore, the shadow of the ball is moving along the ground at -200 ft/s 1 second later.