Question

A military plane maintains an altitude of 15,000 feet over a vast flat desert. It flies at a constant speed on a line that will take it directly over an observer on the ground. At noon, the angle of elevation from the observer’s shoes to the plane is 30 degrees, and the angle of elevation is increasing at a rate of 2.2 degrees per second. Find the speed of the plane at noon. Round off your answer to five significant digits. What is the speed in miles per hour (mph)? Remember that there are 5280 feet in one mile.

The answer is 2303.8 ft/sec or about 1570.8 mph but i don't know how to they arrive at that answer. Please help

Answer 1

As far as I can tell, the time of day makes no difference. So, as usual, draw a diagram.

If the plane is at distance x from the spot directly overhead, its speed is dx/dt, and

tanθ = 15000/x

sec^2θ dθ/dt = -15000/x^2 dx/dt

when θ = 30°, secθ = 2/√3
so, x = 15000/√3 = 8660
2.2°/s = 0.0384 rad/s, so we have

4/3 * 0.0384 = -15000/8660^2 dx/dt
dx/dt = -256 ft/s

I don't like your answer. The angle of elevation is increasing, so the plane is approaching. That means its distance is shrinking, so dx/dt must be negative.

Also, most planes that fly at mach 2 don't do it at 15000 ft.

Answer 2

x = 15000√3 = 25981

4/3 * 0.0384 = -15000/25981^2 dx/dt
dx/dt = -2304 ft/s

My bad.