Question

A projectile launched from the ground with an initial speed of \(20 \mathrm{m} / \mathrm{s}\) and a launch angle \(\theta\) follows a trajectory approximated by $$x=(20 \cos \theta) t, \quad y=-4.9 t^{2}+(20 \sin \theta) t$$ where \(x\) and \(y\) are the horizontal and vertical positions of the projectile relative to the launch point (0,0) a. Graph the trajectory for various values of \(\theta\) in the range \(0<\theta<\pi / 2\) b. Based on your observations, what value of \(\theta\) gives the greatest range (the horizontal distance between the launch and landing points)?

Short answer

Additionally, explain how the trajectory changes for various launch angles in the range \(0 < θ < \frac{\pi}{2}\). Answer: The value of launch angle (θ) that gives the greatest range is \(\frac{\pi}{4}\) or \(45^\circ\). As the launch angle changes within the range \(0 < θ < \frac{\pi}{2}\), the trajectory forms different parabolic shapes. For smaller launch angles, the projectile travels more horizontally with a flatter trajectory, resulting in a shorter time in the air. For larger launch angles, the projectile travels more vertically with a steeper trajectory and takes more time in the air, but it covers a shorter horizontal distance (range). The range is greatest at a launch angle of \(45^\circ\), where the projectile takes an optimal balance between horizontal and vertical displacement.

Step-by-step solution

a. Graphing the trajectory for various values of θ

Here is a step by step process to graph the trajectory for various θ: 1. Choose a number of values of θ in the given range (\(0 < θ < \frac{\pi}{2}\)). These values can be selected according to the desired level of precision (you may start with 5 to 10 equally spaced values of θ). 2. Calculate the launch time for each value of θ by solving \(y(t)\) for each θ for \(y(t) = 0\). To find the corresponding time, use the quadratic formula: $$t = \frac{-20 \sin \theta \pm \sqrt{(20 \sin \theta)^2 - 4(-4.9)(0)}}{2(-4.9)}$$ But as we need the time when projectile lands back, the time must be positive, hence we use only the positive solutions. 3. Now, determine \(x\) and \(y\) coordinates for each θ from \(t=0\) to the time determined above. Plug in selected values to the given equations: $$x(t)=(20\cos\theta)t$$ $$y(t)=-4.9t^2+(20\sin\theta)t$$ 4. Plot the trajectory of the projectile for each θ using the \(x\) and \(y\) coordinates calculated in step 3 on a graph paper or using graphing software. After following the above steps, you would be able to visualize the trajectories of the projectile for various angles of launch (θ).

b. Finding the value of θ that gives the greatest range

Here's how to determine the value of θ which gives the greatest range: 1. Recognize that the range is the horizontal distance between the launch and landing points, which is the value of \(x\) when \(y=0\). We have already found the time it takes for the projectile to return to ground level (\(y=0\)) in part (a). 2. Apply the time to the parametric equation for \(x\): $$R(\theta)=x(t_{\text{ground}})=(20\cos\theta)t_{\text{ground}}$$ 3. Now, to find the value of θ which maximizes the range, we need to find the maximum value of the function \(R(\theta)\). To do this, take the derivative of \(R(\theta)\) with respect to θ and set it equal to 0: $$\frac{d}{d\theta}R(\theta)=0$$ 4. Solve the derivative equation we got above to find the θ that maximizes the range. After solving the equation and using the positive value, you will find that the value of θ that gives the greatest range is: $$\theta = \frac{\pi}{4}$$ So, based on our observations, the value of θ that gives the greatest range is \(\frac{\pi}{4}\), or \(45^\circ\).

Key concepts

Trajectory Analysis

Analyzing the trajectory of a projectile involves understanding its path in motion. This path, also known as the projectile's trajectory, is determined by both horizontal and vertical components. For a projectile launched with an initial speed and angle, its trajectory is influenced by gravity, which acts downwards, affecting only the vertical motion.

In the example given, a projectile is launched at an initial speed of 20 m/s, at various angles within the range of 0 to \(\frac{\pi}{2}\). The projectile's movement can be graphed using two equations:

• The horizontal position: \(x(t) = (20 \cos \theta)t\)
• The vertical position: \(y(t) = -4.9t^2 + (20 \sin \theta)t\)

These equations account for gravity's influence on the projectile's path, while \(\cos \theta\) and \(\sin \theta\) give the directional components. Calculating and plotting these positions for different angles allows us to visualize the projectile's trajectory. By observing these paths, we can understand how different angles affect the reach and final position of the projectile.

Parametric Equations

Parametric equations are critical in modeling projectile motion as they express the coordinates of points on a trajectory as functions of a third parameter, usually time. In our projectile motion case, time \(t\) serves as this parameter.

Using the parametric equations, we can describe the projectile's motion:

• Horizontal position: \(x(t) = (20 \cos \theta)t\)
• Vertical position: \(y(t) = -4.9 t^2 + (20 \sin \theta)t\)

By solving these equations, we can find the x and y coordinates at any given moment. These equations showcase how each component of the projectile's motion is independently influenced by time and gravity.
When you plot these positions for different \(\theta\) values, you can see how altering the angle affects the entire path. Parametric equations, therefore, allow a precise analysis of projectile motion, helping us predict the point of impact and maximum height.

Optimization in Calculus

Optimization in calculus involves finding the maximum or minimum values of a function. In projectile motion, optimizing the launch angle helps in determining the angle that gives the maximum range.

The range R(\(\theta\)) of a projectile is given by the parametric equation for the horizontal position. To optimize for the greatest range, it requires calculating the derivative of the range function and finding when this derivative equals zero.
1. Express range using the launch time when the projectile lands, \(t_{\text{ground}}\), as \(R(\theta)=x(t_{\text{ground}})=(20\cos\theta)t_{\text{ground}}\).2. Differentiate \(R(\theta)\) with respect to \(\theta\) and set the derivative equal to zero to find the critical points.3. Identify the angle that yields the maximum range from these points. For maximum distance, solving this optimization leads to the angle \(\theta = \frac{\pi}{4}\), or 45 degrees. At this angle, both vertical and horizontal components are optimized, achieving the greatest range.