Problem 46 - Calculus Early Transcendentals

Question

If a projectile is fired with an initial velocity of \(v_{0}\) meters per second at an angle \(\alpha\) above the horizontal and air resistance is assumed to be negligible, then its position after \(t\) seconds is given by the parametric equations $$x=\left(v_{0} \cos \alpha\right) t \quad y=\left(v_{0} \sin \alpha\right) t-\frac{1}{2} g t^{2}$$ where \(g\) is the acceleration due to gravity \(\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right).\) (a) If a gun is fired with \(\alpha=30^{\circ}\) and \(v_{0}=500 \mathrm{m} / \mathrm{s},\) when will the bullet hit the ground? How far from the gun will it hit the ground? What is the maximum height reached by the bullet? (b) Use a graphing device to check your answers to part (a). Then graph the path of the projectile for several other values of the angle \(\alpha\) to see where it hits the ground. Summarize your findings. (c) Show that the path is parabolic by eliminating the parameter.

Step-by-Step Solution

Answer

The bullet hits the ground at 51.02 seconds, 22016.88 meters away, reaching a max height of 3193.11 meters. The path is parabolic.

Step 1: Set up the vertical motion equation

Given the equation for vertical motion: \[ y = (v_0 \sin \alpha)t - \frac{1}{2}gt^2 \]we need to find when the projectile hits the ground. This occurs when \(y = 0\).

Step 2: Solve for time of flight

Substitute \(v_0 = 500\,\text{m/s}\), \(\alpha = 30^\circ\), and \(g = 9.8\,\text{m/s}^2\) into the vertical motion equation. The equation becomes:\[ 0 = (500\sin 30^{\circ})t - \frac{1}{2} (9.8)t^2 \]Solve for \(t\): \[ 0 = 250t - 4.9t^2 \]Re-arranging gives:\[ t(250 - 4.9t) = 0 \]Therefore, \(t = 0\) or \(t = \frac{250}{4.9}\). The non-zero solution, \(t = 51.02\,\text{s}\), is the time of flight.

Step 3: Calculate horizontal distance

Use the parametric equation for horizontal motion \(x = (v_0 \cos \alpha)t\). Using the time of flight \(t = 51.02\,\text{s}\):\[ x = (500\cos 30^{\circ}) \times 51.02 \]\[ x = 431.77 \times 51.02 \approx 22016.88\,\text{m} \]

Step 4: Find maximum height

The maximum height is reached at \(t = \frac{v_0 \sin \alpha}{g}\). Calculate this time:\[ t_{max} = \frac{500 \sin 30^{\circ}}{9.8} = \frac{250}{9.8} = 25.51\,\text{s} \]Substitute \(t_{max}\) into the vertical equation:\[ y_{max} = (500 \sin 30^{\circ}) \cdot 25.51 - \frac{1}{2} \cdot 9.8 \cdot (25.51)^2 \]\[ y_{max} = 250 \cdot 25.51 - 4.9 \cdot (25.51)^2 \]\[ y_{max} = 6377.5 - 3184.39 \approx 3193.11\,\text{m} \]

Step 5: Confirm using a graph

Graph the parametric equations using a graphing tool to visualize the projectile's path. Check the values calculated for the time of flight, horizontal distance, and maximum height.

Step 6: Eliminate the parameter to show parabola

To show the path is parabolic, eliminate the parameter \(t\). Solve the horizontal equation:\[ t = \frac{x}{v_0 \cos \alpha} \]Substitute \(t\) into the vertical equation:\[ y = (v_0 \sin \alpha)\frac{x}{v_0 \cos \alpha} - \frac{1}{2}g\left(\frac{x}{v_0 \cos \alpha}\right)^2 \]Simplify:\[ y = x \tan \alpha - \frac{g}{2v_0^2 \cos^2 \alpha}x^2 \]This equation is of the form \(y = ax + bx^2\), which is the equation of a parabola.

Key Concepts

Parametric Equations

In projectile motion, we often use parametric equations to describe the trajectory of an object. Parametric equations allow us to express the position of a projectile in terms of another variable, typically time. In the exercise, the parametric equations used are:

• Horizontal position: \(x = (v_0 \cos \alpha) t\)
• Vertical position: \(y = (v_0 \sin \alpha) t - \frac{1}{2} g t^2\)

These equations depend on initial velocity \(v_0\), launch angle \(\alpha\), and the gravitational acceleration \(g\).
Parametric equations make it straightforward to isolate either the horizontal or vertical component of motion from the overall trajectory. By using time as a parameter, we can simulate various scenarios, observe how factors like launch angle and initial speed affect the projectile's flight, and analyze specific points in its path, such as maximum height and range.

Gravity

Gravity is a crucial force affecting the motion of projectiles. It pulls objects towards the Earth at a constant acceleration. For most calculations involving projectile motion on Earth, gravity is considered to have an acceleration, \(g\), of approximately 9.8 meters per second squared \( \mathrm{m/s^2} \).
In the vertical motion equation \( y = (v_0 \sin \alpha) t - \frac{1}{2} g t^2 \), gravity is represented by the term \( \frac{1}{2} g t^2 \).
This term accounts for the downward pull of gravity, which increases over time, causing the projectile to eventually fall back to the ground.
The effect of gravity is most evident:

• When calculating the maximum height, as gravity slows the projectile's upward vertical motion over time.
• When determining the time of flight, as gravity dictates how quickly the projectile will descend.

Understanding how gravity factors into the parametric equations helps in predicting the real-world behavior of projectiles.

Parabolic Trajectory

A projectile motion without air resistance follows a parabolic trajectory. This trajectory is derived from the parabolic nature of the quadratic equation used in vertical motion. By using the parametric equations provided, the path can be defined precisely and exhibits a symmetric arc.

• The horizontal motion equation \(x = (v_0 \cos \alpha) t\) provides a linear component, as horizontal velocity remains constant.
• The vertical equation \(y = (v_0 \sin \alpha) t - \frac{1}{2} g t^2\) forms a parabola due to the \( g t^2 \) term that influences the symmetrical rise and fall.

To explicitly demonstrate the parabolic nature, we can eliminate the parameter \(t\) from the equations, which leads to a quadratic expression:
\[ y = x \tan \alpha - \frac{g}{2v_0^2 \cos^2 \alpha}x^2 \]
This is a quadratic equation in terms of \(x\), confirming that the projectile's trajectory indeed forms a parabola.

Physics in Calculus

Calculus plays a significant role in physics, especially in understanding projectile motion. The parametric equations employed in the context of projectiles are derived using calculus. Calculus allows us to:

• Find the instantaneous rate of change in the projectile's motion, such as velocity and acceleration.
• Utilize derivatives to calculate velocities. For example, differentiating positions with respect to time gives us horizontal and vertical velocities.
• Use integrals when calculating areas related to motion, though this is less common in basic projectile problems.

Through calculus, we better understand how different elements like initial velocity, launch angle, and gravity shape the trajectory. Determining points such as the maximum height and range involves finding where the derivative (representing change) equals zero, indicating a peak or extreme in the motion.