Answer:
120.9 seconds
Explanation:
*Attached below is a rough sketch of the problem. Point A represents the point where the airplane climbs 30°, BC represents the altitude of the airplane, AC represents the angular displacement of the airplane.
Parameters given:
Angle of elevation = 30°
Speed (angular) of the airplane = 215 ft/sec
Altitude = 13000 ft
We need to find the angular displacement of the airplane to find the time it takes to get to an altitude of 13000 ft. The angular displacement is represented by the hypotenuse of the triangle. Hence, using SOHCAHTOA,
[tex]sin30^{o} = \frac{13000}{hyp} \\\\hyp = \frac{13000}{sin30^{o}}\\ \\hyp = 26000 ft[/tex]
∴ time taken = [tex]\frac{angular displacement}{angular speed}\\ \\[/tex]
[tex]time taken = \frac{26000}{215} \\\\time taken = 120.9 seconds[/tex]
Therefore, it will take the airplane 120.9 seconds to get to an altitude of 13000 ft.
