Question

an airplane is flying east at a constant altitude of 30500 meters. when first seen to the east of an observer, the angle of elevation to the airplane is 63.6 degrees. after 33 seconds, the angle of elevation is 56.1 degrees. find the speed of the airplane

Answer 1

Let h = 30500 m, θ1 = 63.6°, θ2 = 56.1°. If x1 and x2 are the horizontal distances from the observer at the two times, then
tan θ1 = h / x1 and tan θ2 = h / x2,
so x1 = h / tan θ1, x2 = h / tan θ2.

Compute:
x1 = 30500 / tan(63.6°) ≈ 30500 / 2.0145 ≈ 15140 m,
x2 = 30500 / tan(56.1°) ≈ 30500 / 1.4882 ≈ 20496 m.

The plane travels x2 − x1 ≈ 20496 − 15140 = 5356 m in 33 s, so its speed is
v ≈ 5356 / 33 ≈ 162.3 m/s ≈ 584 km/h.

Answer: about 162 m/s (≈ 584 km/h).