A plane is flying horizontally at an altitude of 21000 feet with a ground speed of 875 feet per second. the plane passes directly over an observation tower at sea level. a spectator in the observation tower looks straight up to see it flying over but then lowers their gaze to follow the plane. at what rate is the angle of elevation of the spectator's line of sight changing 8 seconds after the plane has passed overhead?

the angle is changing at a rate of 0.0375 rad/s when the time is t=8 seconds

Step-by-step explanation:

since the line D that starts from the spectator and follows the plane has the following equation

D² = x² + H² ,  where H= altitude of the plane , x= horizontal distance

then for x=v*t , where v=speed of the plane and t=time since the plane has passed overhead , we have for the elevation angle

tan θ = x/y =  v*t /H

θ = tan⁻¹ ( v*t /H)

the rate of change in the angle of the spectator's sight θ with the time is

dθ/dt = 1/1+(v*t /H)² * (v/H) = (v/H)/[1+ (v/H)²*t²]

for t=8 seconds

dθ/dt = (875 ft/s/21000 ft)/[1+ (875 ft/s/21000 ft)²*(8 s)²] = 0.0375 rad/s

therefore the angle is changing at a rate of 0.0375 rad/s when the time is t=8 seconds

answer: idk

step-by-step explanation: