An aircrafts executes a horizontal loop at a speed of $720 k m / h$ with its wings banked at $15^{o}$ . What is the radius of the loop?

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Solution

Consider the problem As the banking of angle is given as

$tan θ = \frac{v^{2}}{r g}$

$r = \frac{v^{2}}{g tan θ}$

And, given

$v = 720 m / h = \frac{720}{3600} = 0.2 m / s$
$g = 9.8 m / s$
$θ = 15^{\circ}$

Therefore,
$r = \frac{(0.2)^{2}}{9.8 \times t a n 15^{\circ}}$
$r = \frac{{(0.2)}^{2}}{9.8 \times 0.26}$
$r = 0.015 m$

Hence, $0.015 m$

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An aircrafts executes a horizontal loop at a speed of 720km/h with its wings banked at 15o. What is the radius of the loop?

Consider the problem As the banking of angle is given as tanθ=v2rgr=v2gtanθAnd, givenv=720m/h=7203600=0.2m/sg=9.8m/sθ=15∘Therefore,r=(0.2)29.8×tan15∘r=(0.2)29.8×0.26r=0.015mHence, 0.015m

An aircrafts executes a horizontal loop at a speed of $720 k m / h$ with its wings banked at $15^{o}$ . What is the radius of the loop?