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As your jet plane speeds down the runway on take off, you decide to determine its acceleration, so you take out your yo-yo and note that when you suspend it, the string makes an angle of 22 degrees with the vertical. What is the acceleration of the plane?

Respuesta :

Answer:[tex]a=3.95 m/s^2[/tex]

Explanation:

Given

string makes an angle of [tex]22^{\circ}[/tex]

Let T be the Tension in the string

From FBD

[tex]Tcos\theta =mg[/tex]----1

[tex]Tsin\theta =ma[/tex]-----2

Divide 1 & 2

[tex]\frac{sin\theta }{cos\theta }=\frac{a}{g}[/tex]

[tex]tan\theta =\frac{a}{g}[/tex]

[tex]a=3.95 m/s^2[/tex]

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