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November 28, 2014

November 28, 2014

Posted by **Teri** on Friday, June 8, 2012 at 10:04am.

- Math -
**Steve**, Friday, June 8, 2012 at 10:35amwhen the ball is at height h, the shadow is at distance x from the base of the light pole. Using similar triangles,

x/50 = (x-30)/h

h = 50(x-30)/x = 50 - 1500/x

now, h = 50-s = 50-16t^2, so

dh/dt = -32t, and

h(1/2) = 50 - 16(1/4) = 46

x/50 = (x-30)/46, so x=375

-32t = 1500/x^2 dx/dt

at t = 1/2,

-16 = 1500/(140625) dx/dt

dx/dt = -1500

so, the shadow is moving toward the pole at 1500 ft/s

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