. A 50.0-kg 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. (a) If the plane’s speed at the lowest point of the circle is 95.0 m/s, what is the minimum radius of the circle so that the acceleration at this point will not exceed 4.00 g? (b) What is the apparent weight of the pilot at the lowest point of the pullout?

Solution 52E A 50.0-kg 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. Given data: Centripetal acceleration at that lowest point is v /r.where v = 95.1 m/s. Acceleration due to gravity is a = 9.81 m/s . Mass of stunt pilot m = 50 kg. (a).The formula to calculate radius of circle is using the equation a = v /r 9.81*4.0 g = (95.1) /r Multiply by r 2 r*9.81 4*0 g = (95.1) Divided by 9.81*4 r = (95.1) /9.81 * r = 230.47 m~230 m. (b).The apparent weight of the pilot at the lowest point of the pullout is given by F = ma(diver) + ma(gravity) F = 50 * *.81 + 50 9.8* F = 2452.4 N.