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An arrow is shot at an angle 32 degree from the horizontal. If it's height is increasing at 200 km/hr, how fast is the arrow itself going?

User Ken Benoit
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Check the picture below.

Bearing in mind that the angle is not changing all the wihle, namely is constant


tan(32^o)=\cfrac{y}{x}\implies x = \cfrac{y}{tan(32^o)}\implies x = \cfrac{1}{tan(32^o)}y \\\\\\ \stackrel{\textit{taking the derivative on bot sides}}{\cfrac{dx}{dt}=\cfrac{1}{tan(32^o)}\cfrac{dy}{dt}}\implies \cfrac{dx}{dt}=\cfrac{1}{tan(32^o)}200\implies \cfrac{dx}{dt}\approx 320.07~(km)/(hr)

An arrow is shot at an angle 32 degree from the horizontal. If it's height is increasing-example-1
User John Flinchbaugh
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