Problem 58 - Calculus Early Transcendentals: Pearson New International Edition

Question

58\. The path of a projectile fired from level ground with a speed of \(v_{0}\) feet per second at an angle \(\alpha\) with the ground is given by the parametric equations $$ x=\left(v_{0} \cos \alpha\right) t, \quad y=-16 t^{2}+\left(v_{0} \sin \alpha\right) t $$ (a) Show that the path is a parabola. (b) Find the time of flight. (c) Show that the range (horizontal distance traveled) is \(\left(v_{0}^{2} / 32\right) \sin 2 \alpha\) (d) For a given \(v_{0}\), what value of \(\alpha\) gives the largest possible range?

Step-by-Step Solution

Answer

(a) Parabolic path; (b) \(\frac{v_0 \sin \alpha}{16}\); (c) Range is \(\frac{v_0^2 \sin 2\alpha}{32}\); (d) Max range at \(45^\circ\).

Step 1: Identify Equation Type

We know that the path of a projectile can be described by parametric equations. For this exercise, the given parametric equations are:\[ x = (v_0 \cos \alpha) t \]\[ y = -16t^2 + (v_0 \sin \alpha) t \]The next step is to eliminate the parameter \(t\) to check the equation type in Cartesian form.

Step 2: Eliminate Parameter to Find the Trajectory Path

To eliminate \(t\), express \(t\) from the \(x\) equation:\[ t = \frac{x}{v_0 \cos \alpha} \]Substitute \(t\) in the \(y\) equation:\[ y = -16 \left(\frac{x}{v_0 \cos \alpha}\right)^2 + (v_0 \sin \alpha) \left(\frac{x}{v_0 \cos \alpha}\right) \]Simplifying gives:\[ y = -\frac{16x^2}{v_0^2 \cos^2 \alpha} + x \tan \alpha \]This is a quadratic equation in terms of \(x\), confirming that the trajectory is a parabola.

Step 3: Calculate Time of Flight

The time of flight occurs when the projectile returns to ground level, thus solving for \(y = 0\):\[ -16t^2 + (v_0 \sin \alpha) t = 0 \]Factor out \(t\):\[ t(-16t + v_0 \sin \alpha) = 0 \]This yields solutions \(t = 0\) or \(t = \frac{v_0 \sin \alpha}{16}\).The time of flight, excluding when \(t = 0\), is \(\frac{v_0 \sin \alpha}{16}\).

Step 4: Derive Projectile Range Formula

The range \(R\) is the horizontal distance traveled, calculated at the time of flight. Substitute \(t = \frac{v_0 \sin \alpha}{16}\) into the \(x\) equation:\[ R = (v_0 \cos \alpha) \left(\frac{v_0 \sin \alpha}{16}\right) \]This simplifies to:\[ R = \frac{v_0^2 \sin 2\alpha}{32} \]

Step 5: Find Angle for Maximum Range

The range formula \( R = \frac{v_0^2 \sin 2\alpha}{32} \) depends on \(\sin 2\alpha\). The maximum value of \(\sin 2\alpha\) is 1, occurring when \(2\alpha = \frac{\pi}{2}\) (i.e., \(\alpha = \frac{\pi}{4}\) radians or 45 degrees). Therefore, the maximum range is achieved when \(\alpha = 45\) degrees.

Key Concepts

Parametric Equations

Parametric equations are a way of expressing the coordinates of points that make up a shape or path. For projectile motion, these equations describe the position of a projectile at any given time. In this exercise, the parametric equations are given as:

• \(x = (v_0 \cos \alpha) t\)
• \(y = -16t^2 + (v_0 \sin \alpha) t\)

These equations represent the horizontal (\(x\)) and vertical (\(y\)) positions of the projectile over time, \(t\). Here, \(v_0\) is the initial velocity, and \(\alpha\) is the launch angle. To understand the trajectory, we manipulate these formulas to remove the time variable \(t\). This involves solving the \(x\)-equation for \(t\) and substituting it into the \(y\)-equation. This turns the trajectory into a quadratic equation in \(x\), revealing a parabolic path. The conversion from parametric to Cartesian form helps us see this clearly.

Quadratic Equations

Quadratic equations frequently appear in physics, especially in problems involving projectile motion. In this context, converting the parametric equations eliminates the parameter \(t\), and we are left with an equation of the form:\[ y = -\frac{16x^2}{v_0^2 \cos^2 \alpha} + x \tan \alpha \]This equation confirms that the path is a parabola, a type of curve defined by a quadratic equation.The negative coefficient of \(x^2\) indicates the parabolic curve opens downward, which is typical for projectile motion due to gravity. Understanding quadratic equations helps in predicting the projectile's path, identifying the vertex (peak point), and calculating values such as the time of flight and range from the motion's apex to the ground.

Time of Flight

The time of flight of a projectile is the total time the projectile spends in the air from launch until it returns to the ground. It is determined by setting the vertical displacement equation to zero, which corresponds to the projectile being at ground level: \[ -16t^2 + (v_0 \sin \alpha) t = 0 \]Solving this quadratic equation gives the time values at which the projectile is at the ground level.The possible solutions are \(t = 0\), which is the initial launch time, and \(t = \frac{v_0 \sin \alpha}{16}\), which is the time of flight. The latter value is particularly important as it tells us how long the projectile stays airborne before landing back on the ground. This information is crucial for planning the launch and predicting where the projectile will land.

Maximum Range

The maximum range of a projectile is the farthest horizontal distance it can reach. This requires optimizing the launch angle \(\alpha\) to achieve the greatest possible distance. By evaluating the range formula:\[ R = \frac{v_0^2 \sin 2\alpha}{32} \]we need to maximize the value of \(\sin 2\alpha\).The function \(\sin 2\alpha\) reaches its maximum value of 1 when \(2\alpha = \frac{\pi}{2}\), or \(\alpha = \frac{\pi}{4}\) radians (45 degrees). Thus, launching at a 45-degree angle will result in the maximum range for a given initial velocity \(v_0\).This is a key insight for understanding how angles affect projectile range, allowing us to strategically choose a launch angle for a desired application in sports, military, or engineering scenarios.