A fighter pilot dives his plane toward the ground at 230

m/s. He pulls out of the dive on a vertical circle. What is the

minimunradius of the circle, so that the normal force exerted on

the pilotby his seat never exceeds three times his weight?

r > = 1799.3 m

or 1799.3 m minimum.

Explanation:

As the pilot cannot pull more than 3 g's, the centripetal acceleration must be less than or equal to three times the force of gravity which is

3*9.8 m/s² = 29.4 m/s².

Acceleration (Centripetal acceleration) is the tangential speed squared divided by the radius of the circle or V^2/r. So we have

Centripetal acceleration=V²/r

230²/r=3*9.8

52900/r=29.4

52900=29.4r

52900/29.4=r

Now

1799.3<=r

r > = 1799.3 m

or 1799.3 m minimum.