What is the angle of Climb (deg) for airplane at 250 KTAS with a climb rate of 4,000 ft/min?

Question

 Aerodynamics question, but related to physics. Any help would be appreciated. Thank you

Answer 1

This can be determined with some conversion of units and trigonometry.

 

250 kph = 287.69 mph

4000 f/min = 45.45 mph

 

The angle of climb, β, is such that

tanβ = 45.45/287.69

tanβ ≈ 0.16

tan^-10.16 ≈ β

9.09° ≈ β

 

The angle of climb is 9.09 degrees.