Related Rates: Object Drop from 200ft Tower, 2 Sec Shadow Rate Calculation

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An object is dropped from rest from a height of 200 ft, 300 feet horizontally across from a 200 ft tall light tower. The object's height above the ground at any given time, t, in seconds, is h= 200 - 16t^2 feet. Exactly 2 seconds after it is dropped, what is the rate at which the shadow is moving across the ground?
 
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i don't even know what to start with, any suggestions?
 
Yes- start by drawing a picture. Draw a vertical line representing the light tower and a horizontal line representing the ground. Mark a point representing the falling object. Draw a straight line from the top of the light tower through the object to the ground. The point where that line touches the ground is the shadow of the object. Finally draw a vertical line from the object to the ground. Now look at it!

You should see two similar triangles so you can set up an equation involving information you know and the distance from the light tower to the shadow. Differentiate that distance with respect to time to get the rate at which it is moving.
 
ill try that
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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