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Question: An aircraft executes a horizontal loop at a speed of 720 km h<sup>-1</sup> with its wings banked at ...

An aircraft executes a horizontal loop at a speed of 720 km h-1 with its wings banked at 15∘15 ^ { \circ } . What is the radius of the loop? (Take g=10 m s−2,tan⁡15∘=0.27\mathrm { g } = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 } , \tan 15 ^ { \circ } = 0.27 )

A

14.8 km

B

14.8 m

C

29.6 km

D

29.6 m

Answer

14.8 km

Explanation

Solution

here

v=720kmh−1=720×518 ms−1=200 ms−1v = 720 \mathrm { kmh } ^ { - 1 } = 720 \times \frac { 5 } { 18 } \mathrm {~ms} ^ { - 1 } = 200 \mathrm {~ms} ^ { - 1 }

θ=15∘,g=10 ms−2\theta = 15 ^ { \circ } , g = 10 \mathrm {~ms} ^ { - 2 }

As tan⁡θ=v2rg\tan \theta = \frac { v ^ { 2 } } { r g }

∴r=v2tan⁡θg=(200 ms−1)2tan⁡15∘×10 ms−2=14815 m=14.8 km\therefore r = \frac { v ^ { 2 } } { \tan \theta g } = \frac { \left( 200 \mathrm {~ms} ^ { - 1 } \right) ^ { 2 } } { \tan 15 ^ { \circ } \times 10 \mathrm {~ms} ^ { - 2 } } = 14815 \mathrm {~m} = 14.8 \mathrm {~km}