Problem 29 - Single Variable Calculus

Question

A projectile is fired from a cannon located on a horizontal plane. If we think of the cannon as being located at the origin \(O\) of an \(x y\) -coordinate system, then the path of the projectile is $$y=\sqrt{3} x-\frac{x^{2}}{400}$$ where \(x\) and \(y\) are measured in feet. a. Find the value of \(\theta\) (the angle of elevation of the gun). b. At what point on the trajectory is the projectile traveling parallel to the ground? c. What is the maximum height attained by the projectile? d. What is the range of the projectile (the distance \(O A\) along the \(x\) -axis)? e. At what angle with respect to the \(x\) -axis does the projectile hit the ground?

Step-by-Step Solution

Answer

a. The angle of elevation of the gun is \(60^{\circ}\). b. The projectile is parallel to the ground at point \((200\sqrt{3}, 300)\). c. The maximum height of the projectile is \(300\) feet. d. The range of the projectile is \(400\sqrt{3}\) feet. e. The projectile hits the ground at an angle of \(-60^{\circ}\).

Step 1: Find the derivative of the given function

First, differentiate the given function \(y = \sqrt{3}x-\frac{x^2}{400}\) with respect to \(x\). Use the power rule and chain rule: \(y' = \frac{d}{dx}(\sqrt{3}x-\frac{x^2}{400})= \sqrt{3}-\frac{2x}{400}\).

Step 2: Find the angle of elevation

To find the angle \(\theta\) of elevation of the gun, evaluate the derivative at \(x=0\). This will give you the tangential slope at that point. Then, use the `tan` inverse function to find the angle: \(y'(0) = \sqrt{3} - \frac{2 * 0}{400} = \sqrt{3}\). Now find \(\theta\): \(\theta = \tan^{-1}(y'(0)) = \tan^{-1}(\sqrt{3}) = 60^{\circ}\).

Step 3: Find the point where the projectile is parallel to the ground

To find the point where the projectile is parallel to the ground, we need to find the point when \(y'(x) = 0\). This will give us the vertex of the parabola, which is also the highest point of the trajectory. Set \(y'(x) = 0\) and solve for \(x\): \(\sqrt{3} - \frac{2x}{400} = 0\). Solve for x: \(x = 200\sqrt{3}\). Now, find the corresponding \(y\) coordinate using the given function: \(y(200\sqrt{3}) = \sqrt{3}(200\sqrt{3})-\frac{(200\sqrt{3})^2}{400} = 300\). So the projectile is parallel to the ground at point \((200\sqrt{3}, 300)\).

Step 4: Find the maximum height of the projectile

We found the highest point of the trajectory in step 3: \((200\sqrt{3}, 300)\). Therefore, the maximum height attained by the projectile is \(300\) feet.

Step 5: Find the range of the projectile

To find the range of the projectile (distance OA along the \(x\) -axis), we need to find the \(x\)-coordinate when the projectile hits the ground, which is when \(y = 0\). Using the given equation, set \(y = 0\) and solve for \(x\): \(\sqrt{3}x-\frac{x^2}{400} = 0\). Factor \(x\) out of the equation: \(x(\sqrt{3}-\frac{x}{400})=0\). So, \(x = 0\) or \(x = 400\sqrt{3}\). The range OA is \(400\sqrt{3}\) feet.

Step 6: Find the angle at which the projectile hits the ground

To find the angle at which the projectile hits the ground, we need to evaluate the derivative at \(x = 400\sqrt{3}\). \(y'(400\sqrt{3}) = \sqrt{3} - \frac{2 * 400\sqrt{3}}{400} = -\sqrt{3}\). Now find the angle: \(\alpha = \tan^{-1}(y'(400\sqrt{3})) = \tan^{-1}(-\sqrt{3}) = -60^{\circ}\). So, the projectile hits the ground at an angle of \(-60^{\circ}\). To summarize the results: a. The angle of elevation of the gun is \(60^{\circ}\). b. The projectile is parallel to the ground at point \((200\sqrt{3}, 300)\). c. The maximum height of the projectile is \(300\) feet. d. The range of the projectile is \(400\sqrt{3}\) feet. e. The projectile hits the ground at an angle of \(-60^{\circ}\).

Key Concepts

Angle of Elevation

The angle of elevation is a critical concept in projectile motion, referring to the angle between the horizontal plane and the initial launch direction of a projectile. In educational terms, it provides the tilt or the initial incline that sets the course for a projectile's journey. For instance, when a cannon fires a projectile, the angle of its barrel with respect to the ground directly impacts the trajectory.
To calculate the angle of elevation in our problem, we take the derivative of the position function, which gives us the slope of the tangent to the curve at any point. At the origin, this slope is equal to the tangent of the angle of elevation. Calculating it yields a result of \(60^\circ\), indicating that the projectile was launched at an angle of \(60^\circ\) above the horizontal. Understanding this fundamental concept is essential for students to predict and analyze the behavior of a projectile in motion.

Trajectory of a Projectile

The trajectory of a projectile describes the path taken by the object through the air, often resembling a parabola in ideal conditions without air resistance. In calculus, we can define this path with an equation in terms of \(x\) and \(y\) variables, representing the horizontal and vertical components, respectively.
By understanding the trajectory equation \(y = \sqrt{3} x - \frac{x^2}{400}\), we can deduce the behavior of the projectile over time. The equation comprises a linear term and a quadratic term, with the quadratic term causing the path to curve downward due to the influence of gravity. Students should appreciate that the highest point of this curve (vertex of the parabola) indicates the apex of the projectile's flight.

Maximum Height of Projectile

When studying maximum height in projectile motion, we are concerned with the highest vertical distance reached by the projectile. This occurs at the vertex of the trajectory's parabolic path. Calculus offers a smooth way to find this: we first locate the horizontal position of the maximum height by setting the derivative of the trajectory function to zero, and then we substitute this value back into the original function.
Through step-by-step analysis, we found that the projectile achieved a maximum height of \(300\) feet. This occurs when the projectile briefly travels parallel to the ground, which happens at the vertex we determined previously. For students, understanding the calculus involved in these calculations is key to mastering projectile motion problems.

Range of Projectile

The range of a projectile is the total horizontal distance the projectile covers from launch to landing. In our exercise, the range is the distance along the \(x\)-axis from the origin to the point where the projectile returns to ground level, depicted by when \(y = 0\).
Solving the projectile's equation for \(y = 0\), we find two solutions for \(x\), one of which is the point of origin \(0\) and the other represents the landing point. In this case, the projectile's range came out to be \(400\sqrt{3}\) feet. This figure is essential for students to envision how far a projectile travels and to plan accordingly in practical scenarios, such as in engineering or ballistics.