Question

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?

Answer 1

Answer:
`a=3.95 m/s^2`

Step-by-step explanation:

Given

string makes an angle of
`22^(\circ)`

Let T be the Tension in the string

From FBD

`Tcos\theta =mg`----1

`Tsin\theta =ma`-----2

Divide 1 & 2

`(sin\theta )/(cos\theta )=(a)/(g)`

`tan\theta =(a)/(g)`

`a=3.95 m/s^2`