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 lightâ€™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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