Question

B) A skilled jet fighter flies a stunt plane in a vertical circle of 1200 ft

Answer 1

The jet fighter's acceleration at the highest point is 17.56 ft/s²

The jet fighter's acceleration at the lowest point is 35.12 ft/s²

The given parameters;

radius of the circular path, r = 1200 ft

constant velocity, v = 140 mi/h = 205.3 ft/s

The formula for centripetal acceleration in a circular path is given as;

$a_c = \frac{v^2}{r}$

The jet fighter's acceleration at the highest point is calculated as;

h = 2r = 2 x 1200 ft = 2,400 ft

$a_c = \frac{v^2}{r} \\\\a_c = \frac{(205.3)^2}{2400} \\\\a_c = 17.56 \ ft/s^2$

The jet fighter's acceleration at the lowest point is calculated as;

h = r = 1,200 ft

$a_c = \frac{v^2}{r} \\\\a_c = \frac{(205.3)^2}{1200} \\\\a_c = 35.12\ ft/s^2$

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