How Is a Pilot's Apparent Weight Calculated During a Vertical Dive Pullout?

[raiderIV]

Homework Statement

A 50.0kg stunt pilot who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane.

If the plane's speed at the lowest point of the circle is 95.0m/s, what should the minimum radius of the circle be in order for the acceleration at this point not to exceed 3.00 g?
Correct Answer for that was 307m

Question i need help on is:
What is the apparent weight(in Newtons) of the pilot at the lowest point of the pullout?

Homework Equations

W = mg
g=v^2/r
1g = 9.81

The Attempt at a Solution

I assumed that the easy solution would be to find the weight at 3g's so:
50*3*9.81 = 1471.5

Wasn't accepted as correct answer so i decided to solve it like:
F = ma = m(v^2)/r
50*(95^2)/307 = 1469.9

Also, not taken as the correct answer.


Please let me know what I am missing here
Thanks,
~John

 

[flatmaster]

the acceleration of the plane is 3g, but did Earth's gravity turn off?

 

[raiderIV]

the acceleration of the plane is 3g, but did Earth's gravity turn off?

Ok ok! Its ok everyone! I turned the gravity back on! ^_^

Thank you for the help. I didnt think to add that in there and it worked beautifully
~John