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ok... here's the problem.

A rocket is launched vertically from a point on the ground that is 100 horizontal meters from an observer with binoculars. The rocket is rising vertically and its height above the ground (in meters) is given by : [tex] y(t)=60t-5t^2 [/tex]

Two seconds after launch, how fast must the observer change the angle of elevation of her line of sight to keep the rocket in the binoculars?

I drew out the figure and...i know that i need to somehow compare the rate of change of the height of the rocket and the angle of her line of sight. So the rate of change of the height is y ' , which i got [tex] y' = t - 10t[/tex]

now how can i relate the two? I can plug in 2 seconds in y' to get the rate at wich the height is changing, but what about the angle? as always, any help would be awesome.

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# Angle of line of sight for a vertically lauched rocket

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