Problem 44 - College Trigonometry

Question

Eliminate the parameter in the equations for projectile motion to show that the path of the projectile follows the curve $$ y=-\frac{g \sec ^{2}(\theta)}{2 v_{0}^{2}} x^{2}+\tan (\theta) x+s_{0} $$ Use the vertex formula (Equation 2.4) to show the maximum height of the projectile is $$ y=\frac{v_{0}^{2} \sin ^{2}(\theta)}{2 g}+s_{0} \quad \text { when } \quad x=\frac{v_{0}^{2} \sin (2 \theta)}{2 g} $$

Step-by-Step Solution

Answer

The projectile path is a parabola: \( y = -\frac{g \sec^2(\theta)}{2 v_0^2} x^2 + \tan(\theta) x + s_0 \). The maximum height is \( y = \frac{v_0^2 \sin^2(\theta)}{2g} + s_0 \).

Step 1: Write the Parametric Equations

In projectile motion, the parametric equations for the position of the projectile at time \(t\) are given by: \[ x = v_0 \cos(\theta) t \] and \[ y = s_0 + v_0 \sin(\theta) t - \frac{1}{2} g t^2 \]where:- \(x\) and \(y\) are the horizontal and vertical positions, respectively.- \(v_0\) is the initial velocity.- \(\theta\) is the angle of projection.- \(s_0\) is the initial vertical position.- \(g\) is the acceleration due to gravity.

Step 2: Solve for Time from x Equation

To eliminate the parameter \( t \), solve for \( t \) in terms of \( x \):\[ t = \frac{x}{v_0 \cos(\theta)} \]

Step 3: Substitute into y Equation

Substitute the expression for \( t \) from Step 2 into the equation for \( y \):\[ y = s_0 + v_0 \sin(\theta) \left( \frac{x}{v_0 \cos(\theta)} \right) - \frac{1}{2} g \left( \frac{x}{v_0 \cos(\theta)} \right)^2 \]Simplify to obtain:\[ y = s_0 + x \tan (\theta) - \frac{g}{2 v_0^2 \cos^2(\theta)} x^2 \]Recognize that \( \sec^2(\theta) = \frac{1}{\cos^2(\theta)} \), leading to the projectile path:\[ y = -\frac{g \sec^2(\theta)}{2 v_0^2} x^2 + \tan(\theta) x + s_0 \]

Step 4: Determine the Maximum Height

The maximum height is achieved at the vertex of the parabolic path described by:\[ y = -\frac{g \sec^2(\theta)}{2 v_0^2} x^2 + \tan(\theta) x + s_0 \]The vertex \(x\)-coordinate is:\[ x = -\frac{b}{2a} \]where \( a = -\frac{g \sec^2(\theta)}{2 v_0^2} \) and \( b = \tan(\theta) \), so:\[ x = \frac{v_0^2 \sin (2\theta)}{2g} \]

Step 5: Substitute x Into y for Maximum Height

Substitute the value of \( x \) from Step 4 into the parabolic equation:\[ y = -\frac{g \sec^2(\theta)}{2 v_0^2} \left(\frac{v_0^2 \sin(2\theta)}{2g}\right)^2 + \tan(\theta) \left(\frac{v_0^2 \sin(2\theta)}{2g}\right) + s_0 \]Simplifying gives the maximum height:\[ y = \frac{v_0^2 \sin^2(\theta)}{2g} + s_0 \]

Step 6: Confirm Maximum Height and Coordinates

We have confirmed that the projectile reaches its maximum height \( y = \frac{v_0^2 \sin^2(\theta)}{2g} + s_0 \) when \( x = \frac{v_0^2 \sin(2\theta)}{2g} \). This implies the vertex point on the projectile path is correct and coincides with the maximum height formula.

Key Concepts

Parametric Equations

Parametric equations are a way to represent the path of an object moving along a curve by expressing its coordinates as functions of a parameter, usually time \(t\).
In projectile motion, the motion of an object thrown or projected into the air can be described using two parametric equations:

• The horizontal position is given by: \( x = v_0 \, \cos(\theta) \, t \), where \(v_0\) is the initial velocity and \(\theta\) is the angle of projection.
• The vertical position is given by: \( y = s_0 + v_0 \, \sin(\theta) \, t - \frac{1}{2} \, g \, t^2 \), where \(s_0\) is the initial vertical position and \(g\) is the acceleration due to gravity (usually \(9.8 \, \text{m/s}^2\)).

These equations help analyze various attributes of the projectile's motion, such as its path, velocity, and time-related properties.
To simplify or further manipulate these equations, components like the elimination of \(t\) are often applied to derive additional insights such as the overall path the projectile takes.

Vertex Formula

The vertex formula is crucial when dealing with parabolic paths, which are common in projectile motion scenarios.
In general, for a quadratic equation of the form \(y = ax^2 + bx + c\), the vertex provides the point at which the parabola reaches either its highest or lowest point.

• The x-coordinate of the vertex is calculated by \(x = -\frac{b}{2a}\).

This x-coordinate helps find the point of maximum height for the projectile by substituting it back into the equation.
In our scenario, the congruent vertex calculation gives the x-coordinate where the projectile achieves maximum height:
\[x = \frac{v_0^2 \, \sin(2\theta)}{2g} \] Once the x-coordinate is determined, substitute into the quadratic path equation to calculate the actual maximum height.

Maximum Height

The maximum height in projectile motion is the highest vertical position reached by the projectile.
To find it, we use the vertex form of the parabolic path equation. Calculating the vertex for the path, which involves knowing where \(x = \frac{v_0^2 \, \sin(2\theta)}{2g}\), we substitute this x value into the path equation to find the y-coordinate at that position.
The maximum height can be explicitly calculated with the formula: \[ y = \frac{v_0^2 \, \sin^2(\theta)}{2g} + s_0 \] Here, every term is crucial. \(v_0\) reflects the initial velocity, \(\theta\) is the angle of projection, and \(s_0\) is the initial height.
This specific height is pivotal for understanding how high the projectile can go under the given launch conditions.

Trigonometric Identities

Trigonometric identities are mathematical equations involving trigonometric functions that are true for every value of the parameter.
In projectile motion, trigonometric identities are utilized to simplify and derive paths and heights of projectiles.
Specifically, identities like:

• \(\sec^2(\theta) = 1 + \tan^2(\theta)\), which was used in simplifying the parabolic equation of the projectile's path.
• \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\), used to simplify expressions involving double angles and when determining where maximum height occurs.

These identities are proven tools that make resolving or reconfiguring equations based on angles manageable and essential to finding solutions in projectile scenarios.
Mastering these can clarify many problems that involve varying angles and paths.