Posted by **Linda** on Monday, April 2, 2012 at 4:18am.

A projectile is fired upward 200 ft from an observer. The height in feet of the projectile after t seconds is given by s=-16t2 + 180t. What is the rate of change of the angle of elevation of the observer to the projectile after 7 seconds?

I don't understand how to get there!

- Math -
**Steve**, Monday, April 2, 2012 at 10:59am
if θ is the angle of elevation,

y = -16t^2 + 180t

tanθ = y/200

sec^2θ dθ/dt = 1/200 dy/dt = 1/200 (-32t + 180)

= -32/200 (t-5)

dθ/dt = -4/25 (t-5) cos^2θ

= -4/25 (t-5) (200^2/(200^2 + y^2)

y(7) = -16*49 + 180*7 = 476

at t=7,

dθ/dt = -4/25 * 2 * 40000/266576 = -.048

Seems kinda small, so you better check the details.

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