Question

A pilot performs an evasive maneuver by diving vertically at 270. If he can withstand an acceleration of 9.0 's without blacking out, at what altitude must he begin to pull out of the dive to avoid crashing into the sea? y= ?m

Answer 1

Answer:

The pilot must be began at an altitude of 826.53 m to avoid crash into the sea.

Explanation:

Given that,

Velocity = 270 m/s

Acceleration = 9.0g s

We need to calculate the altitude

Using formula of centripetal acceleration

$a_{c}=\dfrac{v^2}{r}$

$r=\dfrac{v^2}{a_{c}}$

Where, v = velocity

r = altitude

a = acceleration

Put the value into the formula

$r=\dfrac{270^2}{9.0\times9.8}$

$r=826.53\ m$

Hence, The pilot must be began at an altitude of 826.53 m to avoid crash into the sea.