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Chapter 3: Problem 93

An airplane flies horizontally above the flat surface of a desert at an altitude of \(5.00 \mathrm{~km}\) and a speed of \(1000 . \mathrm{km} / \mathrm{h}\) If the airplane is to drop a care package that is supposed to hit a target on the ground, where should the plane be with respect to the target when the package is released? If the target covers a circular area with a diameter of \(50.0 \mathrm{~m}\), what is the "window of opportunity" (or margin of error allowed) for the release time?

Short Answer

Expert verified
The airplane should be 8894 meters horizontally from the target when the package is released. The window of opportunity for the release time is between 31.93 seconds and 32.05 seconds.

Step by step solution

01

Determine the time for the package to hit the ground

Since the airplane is flying horizontally, the only force acting on the package vertically is gravity. We can use the equation of motion: \(h = v_{0_y}t + 0.5at^2\) Where \(h = 5000 \ m\) is the altitude, \(v_{0_y} = 0\) is the initial vertical velocity, \(a = 9.81 \ m/s^2\) is the acceleration due to gravity, and \(t\) is the time in seconds. The equation for the time becomes: \(5000 = 0.5(9.81)t^2\) Solving for \(t\), we get: \(t = \sqrt{\frac{2 * 5000}{9.81}} \approx 31.99 \ s\) Now that we have the time, we can determine the horizontal position of the airplane relative to the target.
02

Calculate the horizontal position of the airplane

To find the horizontal position of the airplane relative to the target when the package should be released, we can use the horizontal component of the motion equation: \(x = v_{0_x}t\) Where \(v_{0_x}\) is the horizontal speed of the airplane in meters per second, and \(x\) is the horizontal distance. First, we need to convert the airplane's horizontal speed from km/h to m/s: \(v_{0_x} = 1000 \frac{km}{h} * \frac{1000 \ m}{1 \ km} * \frac{1 \ h}{3600 \ s} = 277.78 \ m/s\) Now, we can find the horizontal distance: \(x = (277.78 \ m/s)(31.99 \ s) \approx 8894 \ m\) The airplane should be 8894 meters horizontally from the target when the package is released.
03

Determine the window of opportunity

Now we need to find the time intervals during which the care package will still land within the target area. Since the diameter of the circular target area is \(50 \ m\), the package can fall a maximum of 25 meters from the center in any direction. We will use the equation for horizontal distance: \(x = v_{0_x}t\) But instead of using 8894 meters, we will use 8869 meters (25 meters less) and 8919 meters (25 meters more). We will solve for the time in each case: Case 1: \(x = 8869 \ m\) \(t = \frac{8869}{277.78} \approx 31.93 \ s\) Case 2: \(x = 8919 \ m\) \(t = \frac{8919}{277.78} \approx 32.05 \ s\) The window of opportunity for the release time is between 31.93 seconds and 32.05 seconds.

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