Question

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

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

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.

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.

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.

Key concepts

Motion Equations

Motion equations help us describe the movement of an object through space. In the context of projectile motion, which is what we're dealing with when the airplane drops a package, these equations can predict where and when the package will land.
The basic motion equation we use in the vertical direction is:\[ h = v_{0_y}t + 0.5at^2 \]- **Vertical Component**: In our scenario, since the package initially falls from rest vertically, the initial vertical velocity \(v_{0_y}\) is zero. The altitude \(h\) is given as 5000 meters. The only force at play is gravity, with an acceleration \(a\) of \(9.81 \, \text{m/s}^2\). By solving the motion equation, we determine that the package will take approximately 32 seconds to reach the ground.- **Horizontal Component**: Here, we account for the airplane's speed. The horizontal component doesn't get affected by gravity in this scenario since the motion is purely horizontal from the physics perspective. Using the equation \(x = v_{0_x}t\), we translate the plane's speed into a distance covered over time. We found that the plane has to be 8894 meters away from the target at the time of release.

Horizontal Velocity

Horizontal velocity plays a critical role in dropping a package from a moving object such as an airplane. Unlike vertical motion, horizontal motion remains uniform in this example, as there are no notable forces acting sideways, provided we neglect air resistance.
The airplane's horizontal speed, initially given in km/h, needs conversion to meters per second (m/s) for accurate calculation:- **Conversion**: The velocity was provided as 1000 km/h. By converting to m/s using the factor \(1 \text{ km/h} = \frac{1000}{3600} \text{ m/s}\), we obtain \( v_{0_x} = 277.78 \, \text{m/s}\).This constant horizontal speed allows us to use the equation \( x = v_{0_x}t \) to determine how far ahead of the target the airplane must be when releasing the payload. This simple linear relationship tells us that distance is directly proportional to time, as long as the airplane maintains this consistent speed.

Acceleration due to Gravity

The acceleration due to gravity is a fundamental concept in physics. This force pulls objects towards Earth's surface at approximately \(9.81 \, \text{m/s}^2\).
When the airplane releases the package, gravity is the sole force affecting its vertical motion.- **Vertical Motion**: The equation \(h = 0.5at^2\) is used to determine how long it takes for the package to fall from its release point to the ground. Here, \(a\) stands for the gravitational acceleration, and because the initial vertical velocity \(v_{0_y}\) is zero, the equation simplifies significantly.- **Effect on Timing**: Since the altitude of release is 5000 meters, solving the equation results in the time of approximately 32 seconds for the package to hit the ground. This value is crucial for ensuring the package is released at the right moment to reach the target area.Understanding gravity’s influence helps in predicting and planning the release point of falling objects from any height.