Question

A plane diving with constant speed, at an angle of \(49.0^{\circ}\) with the vertical, releases a package at an altitude of \(600 . \mathrm{m} .\) The package hits the ground 3.50 s after release. How far horizontally does the package travel?

Short answer

Answer: The package travels 223.5 meters horizontally before hitting the ground.

Step-by-step solution

Find the vertical and horizontal components of the velocity

The plane is diving at an angle of \(49.0^{\circ}\) to the vertical. We should find the vertical and horizontal components of the speed of the plane to then find how far horizontally the package travels. Let \(v\) be the constant speed of the plane in its dive. We can find the vertical component of the velocity \(v_v\) and horizontal component of the velocity \(v_h\) using trigonometry as: \(v_v = v \sin(49.0^{\circ})\) \(v_h = v \cos(49^{\circ})\) We will use these components in the next steps.

Find the vertical speed component

Given that the package hits the ground 3.50 s after release and the altitude when it's released is 600 m, we can use the time and vertical component of the velocity to find the vertical speed component. Using the equation for the distance fallen, \(h = \frac{1}{2} g t^2\), where \(h\) is the altitude (600 m), \(g\) is the acceleration due to gravity (9.81 \(\mathrm{m/s^2}\)) and \(t\) is the time (3.50 s). Plugging these values, we get: \(600 = \frac{1}{2} (9.81)(3.50)^2\) From this equation, we can find the initial vertical speed, \(v_v\): \(v_v = 94.04\ \mathrm{m/s}\)

Find the horizontal speed component

Now, we can use the vertical speed component and the given diving angle to find the horizontal speed component \(v_h\). Using the trigonometric relation that we mentioned in step 1: \(v_h = \frac{v_v}{\sin(49^{\circ})} \cos(49^{\circ})\) Plugging the vertical speed component \(v_v = 94.04 \ \mathrm{m/s}\), we get: \(v_h = 63.87\ \mathrm{m/s}\)

Calculate the horizontal distance

Finally, we can use the horizontal speed component and the time to find the horizontal distance traveled by the package. The formula to calculate the horizontal distance \(d\) is: \(d = v_h t\) Plugging in the values \(v_h = 63.87\ \mathrm{m/s}\) and \(t = 3.50\ \mathrm{s}\), we get: \(d = (63.87)(3.50) = 223.5\ \mathrm{m}\) Hence, the package travels horizontally for 223.5 meters before hitting the ground.

Key concepts

Trigonometry in Physics

In the realm of physics, and particularly in projectile motion, trigonometry plays an essential role in breaking down complex problems into manageable parts. The exercise in question, involving a plane releasing a package while diving, can be intricately deconstructed using trigonometry. Consider the speed of the plane as a vector that is angled with respect to the vertical. Using trigonometric functions, namely sine and cosine, we can resolve this vector into its vertical and horizontal components.

For the vertical component, which aligns with the gravitational pull, we use the sine function:
\(v_v = v \sin(\theta)\).
For the horizontal component, which is perpendicular to gravity and represents the forward movement, we use the cosine function:
\(v_h = v \cos(\theta)\).

The angle \(\theta\) in our case is 49 degrees. By decomposing the velocity, it becomes straightforward to analyze the motion in each direction independently, allowing us to address the vertical and horizontal movement of the package as separate entities. This application of trigonometry is vital and ubiquitous in physics, providing clarity and simplicity to problems that may initially appear convoluted.

Kinematics Equations

Kinematics, the study of motion without considering forces, gives us the equations needed to quantify and predict the behavior of moving objects. These equations are the toolkit for solving motion-based problems, like the one at hand, where a package is dropped from a diving plane.

The second step in our solution uses a kinematic equation to relate the distance an object falls with the time it takes to fall and the acceleration due to gravity. The equation is:
\(h = \frac{1}{2} g t^2\),
where \(h\) stands for height, \(g\) for gravitational acceleration, and \(t\) for time.

Using this equation, we determined the vertical speed component of the package at the moment of release, critical for later finding out how far the package will travel horizontally. There's another kinematic equation that we use indirectly in this problem:
\(d = vt\),
to find the horizontal distance by multiplying the horizontal speed with the time of travel. The versatile nature of kinematics equations allows us to use them in several contexts, emphasizing the value of mastering these fundamental relations in physics.

Components of Velocity

Velocity is a vector quantity, which means it has both magnitude and direction. Understanding the components of velocity is essential when analyzing projectile motion, as depicted in our textbook scenario. When the plane releases the package, it does so with a certain speed in a specific direction. However, due to gravity, the actual path traversed by the package is not straight but a parabola.

In our solution, we split the plane's dive speed into two perpendicular components: vertical \(v_v\) and horizontal \(v_h\). The vertical component is influenced by gravity and dictates how long the package will take to hit the ground. On the other hand, the horizontal component is consistent throughout the package's flight, assuming no air resistance, and controls how far it will travel before impact.

Interconnection of Components

When the package is in the air, both components of velocity act simultaneously. The horizontal distance covered by the package before it reaches the ground is obtained by multiplying the horizontal velocity component by the time it is airborne, resulting in the formula:
\(d = v_h t\).

This concept underscores the idea that motion in one direction is independent of motion in another direction, a fundamental principle in classical mechanics that allows for the analyze of complex movements in a simplified manner.