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Chapter 3: Problem 69

A jet is diving vertically downward at \(1200 \mathrm{km} / \mathrm{h}\). If the pilot can withstand a maximum acceleration of \(5 g\) (i.e., 5 times Earth's gravitational acceleration) before losing consciousness, at what height must the plane start a quarter turn to pull out of the dive? Assume the speed remains constant.

Short Answer

Expert verified
The plane must start the quarter turn from a height of approximately 2293 meters.

Step by step solution

01

Convert Velocity to m/s

First convert the given velocity from km/h to m/s, using the relation: \(1 \mathrm{km} / \mathrm{h} = 0.2778 \mathrm{m} / \mathrm{s}\). Thus, the velocity \(v = 1200 \mathrm{km} / \mathrm{h} = 333.6 \mathrm{m} / \mathrm{s}\).
02

Convert Acceleration to m/s²

Next, convert the acceleration from \(g\) to \(m/s^2\), using the relation: \(1g = 9.8 m/s^2\). Thus, the acceleration \(a = 5g = 49 m/s^2\).
03

Calculate Radius of the Turn

Using the formula for the radius of circular motion \( r = \frac{v^2}{a} \), where \(v\) is the velocity and \(a\) is the centripetal acceleration, we can find the radius. Substituting our calculated values, we get \( r = \frac{(333.6 \mathrm{m} / \mathrm{s})^2}{49 \mathrm{m} / \mathrm{s}^2} = 2292.95 \mathrm{m}\).
04

Determine Height for Quarter Turn

As the jet needs to start the quarter turn, the height can be considered as an arc representing a quarter of the circle's circumference, which is equivalent to the radius of the circle. Thus, the required height \(h = r = 2292.95 \mathrm{m}\).

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