Flying High with Jet Fighter's Acceleration

Have you ever wondered about the acceleration of a skilled jet fighter flying a stunt plane?

What is the jet fighter's acceleration at the highest point and lowest point while flying in a vertical circle of 1200 ft radius at a constant speed of 140 mi/h?

Calculating Jet Fighter's Acceleration

At the highest point:

The jet fighter's acceleration is 17.56 ft/s².

At the lowest point:

The jet fighter's acceleration is 35.12 ft/s².

Understanding the Jet Fighter's Acceleration

When a skilled jet fighter flies a stunt plane in a vertical circle with a radius of 1200 ft at a constant speed of 140 mi/h, we can calculate the acceleration at the highest and lowest points of the circle.

At the highest point, which is 2400 ft above the ground, the jet fighter's acceleration is 17.56 ft/s². This acceleration is determined by the formula for centripetal acceleration in a circular path: ac = v²/r.

Substituting the values of velocity (v = 205.3 ft/s) and radius (r = 1200 ft) into the formula, we get ac = (205.3)²/2400 = 17.56 ft/s².

At the lowest point, which is 1200 ft above the ground, the jet fighter's acceleration is 35.12 ft/s². Substituting the same values into the formula, we get ac = (205.3)²/1200 = 35.12 ft/s².

These calculations show the difference in acceleration experienced by the jet fighter at different points in the circular path. Despite flying at a constant speed, the acceleration varies due to the changing direction of motion in the vertical circle.