An aircraft executes a horizontal loop at a speed of 720 km - h with its wings banked at 15- - What is the radius of the loop- take tan 15-0-27 and g -10 ms -2

Given speed of aircraft v= 720 km/h = 200 m/s Angle of banking θ = 15^∘ Using relation v^2 = Rg tanθ we have nbsp; R =v^2g tanθ = (200)^210 ×tan 15^∘ = (200 ...

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Standard XII

Physics

Banking Angle

An aircraft e...

Question

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

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Solution

Given speed of aircraft $v = 720 k m / h = 200 m / s$
Angle of banking $θ = 15^{\circ}$
Using relation $v^{2} = Rg tan θ$ we have
$R = \frac{v^{2}}{g tan θ} = \frac{(200)^{2}}{10 \times tan 15^{\circ}} = \frac{(200)^{2}}{10 \times 0.27}$
$= 14814.81 m$
$≅ 14.81 k m$

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An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15∘. What is the radius of the loop? (take tan15∘=0.27) and g=10 ms−2

Given speed of aircraft v=720 km/h=200 m/s Angle of banking θ=15∘ Using relation v2=Rg tanθ we have R=v2g tanθ=(200)210×tan15∘=(200)210×0.27 =14814.81 m ≅14.81 km

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

Given speed of aircraft $v = 720 k m / h = 200 m / s$
Angle of banking $θ = 15^{\circ}$
Using relation $v^{2} = Rg tan θ$ we have
$R = \frac{v^{2}}{g tan θ} = \frac{(200)^{2}}{10 \times tan 15^{\circ}} = \frac{(200)^{2}}{10 \times 0.27}$
$= 14814.81 m$
$≅ 14.81 k m$