Posted by **rolan** on Wednesday, August 8, 2012 at 10:29am.

A street light (light source) is 15 m above the ground. A ball is freely released from a location of 9 m from the light with the same elevation as the light’s. While the ball is falling, it casts a shadow on the ground. Assuming the gravity is 10 m/s2, calculate the speed of the shadow of the ball at 0.5 s after the ball has been released.

- math -
**Steve**, Wednesday, August 8, 2012 at 10:43am
the height y of the ball at time t is

y = 15 - 5t^2

The shadow's position x meters from where the ball will hit, is found by using similar triangles:

y/x = 15/(9+x)

(15 - 5t^2)/x = 15/(9+x)

dx/dt = -(9+x)/t

at t = .5, x = 99, so

dx/dt = -(9+99)/(.5) = -216 m/s

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