What is the apparent weight of the pilot at the lowest point of the pullout in this case?

Question

A stunt pilot of mass 55.0 kg 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 95.8m/s , what is the minimum radius of the circle for the acceleration at this point not to exceed 4.00 g?
Take the free fall acceleration to be = 9.80 .What is the apparent weight of the pilot at the lowest point of the pullout in this case?
Take the free fall acceleration to be = 9.80 m/s^2.

Answer 1

a) The centripital acceleration
a=V^2/R
V - tangential velocity
R- the radius of the tragectory
Then
R= V^2/a and for a=4g max
R= V^2/4g
R=(95.8)^2/(4 x 9.80)=234 m

b) a(lowest point)=a+g=4g+g=5g
W=mat=a(lowest point)=5mg
W= 5 x 55.0 x 9.80=2700N

Answer 2

centripetal acceleration =4.00 g =4 x 9.8=39.2 m/s^2

centripetal acceleration = (speed v )^2 /(radius 'r')

radius 'r'= (speed v )^2 / centripetal acceleration

r = 95.8 x 95.8 /39.2= 234 meter

The radius of circle is 234 meter
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apparent weight = true weight + centripetal force

apparent weight = 55 x9.8 +55x39.2= 2695 N

Apparent weight of the pilot is 2695 N or 275 kg wt