Question

(121 m/s, or 43
4. The pilot of an airplane, which has been diving at a speed of
540 km/h, pulls out of the dive at constant speed.
(a) What is the minimum radius of the plane's circular path in
order that the acceleration of the pilot at the lowest point
will not exceed 7 g?
(b) What force is applied on an 80 kg pilot by the plane seat at
the lowest point of the pull-out? (328 m, 6.3 X 10N)

Answer 1

Answer:

(a) r = 328 m

(b) F = 6271.8 N

Explanation:

speed, v = 540 km/h = 150 m/s

Acceleration, a = 7 g

(a) Let the radius is r.

$a =\frac{v^{2}}{r}\\7\times 9.8 = \frac{150\times 150}{r}\\r = 328 m$

(b) The force on 80 kg pilot at the lowest point is

$F =m \times g + m\frac{v^{2}}{r}\\F = 80\times9.8 +80\times \frac{150\times 150}{328}\\\\F =784 + 5487.8 = 6271.8 N$