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A hemispherical dome has a diameter of 100m. A spotlight is placed on point B along the circumference on ground level. A balloon is released at point A on the center at a rate of 2m/s. how fast is the shadow of the balloon move on the roof when the balloon is 25m high?

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There will be no shadow because (i) the balloon is not high enough for the spotlight to reach it (the dome is in the way), and (ii) if the spotlight is on the ground, then it will be casting a shadow of the balloon up into space, not down onto the roof. – Théophile Aug 22 '12 at 2:31
@Théophile I think point A is on the ground and at the center of the big circle? And we ignore diffraction. – FrenzY DT. Aug 22 '12 at 2:34
Ah, the balloon is inside the dome? And so is the spotlight? I had understood that the balloon was released from the top of the roof. A clearer description of the problem would help! (The picture below helps a lot ...) – Théophile Aug 22 '12 at 2:48

Sketch in GeoGebra

Click here for higher resolution.

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This is a partial answer, meant to providing HINT only. Next, differentiate the arc length with respect to $t$, and you have the velocity. – FrenzY DT. Aug 22 '12 at 2:52
how did you make that graph! I would like to learn how. – yiyi Oct 26 '12 at 2:11
@MaoYiyi I made that using the free open source software called GeoGebra. You may find that link in the community ads, i.e. here. – FrenzY DT. Oct 26 '12 at 2:14

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