An aircraft executes a horizontal loop at a speed of $720 k m h^{- 1}$ with its wings banked at $15^{0}$ . What is the radius of the loop?
(Take $g = 10 m s^{- 2}, t a n 15^{0} = 0.27$ )

14.8 km

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14.8 m

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29.6 km

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29.6 m

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Solution

The correct option is A 14.8 km
Here,
$v = 720 k m h^{- 1} = 720 \times \frac{5}{18} m s^{- 1}$
$θ = 15^{0}, g = 10 m s^{- 2}$

As $t a n θ = \frac{v^{2}}{r g}$

$∴ r = \frac{v^{2}}{t a n θ g} = \frac{(200 m s^{- 1})^{2}}{t a n 15^{0} \times 10 m s^{- 2}}$

$= 14815 m = 14.8 k m$

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Q. An aircraft executes a horizontal loop at a speed of

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An aircraft executes a horizontal loop with a speed of 150 m/s with its, wings banked at an angle of $120^{\circ}$ . The radius of the loop is $(g = 10 m / s^{2})$

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720 k m / h

with its wings banked at

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An aircraft executes a horizontal loop at a speed of 720kmh−1 with its wings banked at 150. What is the radius of the loop?(Takeg=10ms−2,tan150=0.27)

The correct option is A 14.8 kmHere,v=720kmh−1=720×518ms−1θ=150,g=10ms−2 As tanθ=v2rg∴r=v2tanθg=(200ms−1)2tan150×10ms−2=14815m=14.8km

An aircraft executes a horizontal loop at a speed of $720 k m h^{- 1}$ with its wings banked at $15^{0}$ . What is the radius of the loop?
(Take $g = 10 m s^{- 2}, t a n 15^{0} = 0.27$ )

The correct option is A 14.8 km
Here,
$v = 720 k m h^{- 1} = 720 \times \frac{5}{18} m s^{- 1}$
$θ = 15^{0}, g = 10 m s^{- 2}$

As $t a n θ = \frac{v^{2}}{r g}$

$∴ r = \frac{v^{2}}{t a n θ g} = \frac{(200 m s^{- 1})^{2}}{t a n 15^{0} \times 10 m s^{- 2}}$

$= 14815 m = 14.8 k m$