Posted by Linda on .
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!
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
dθ/dt = -4/25 * 2 * 40000/266576 = -.048
Seems kinda small, so you better check the details.