Problem 41 - Algebra and Trigonometry

Question

If a projectile is fired with an initial speed of \(v_{0}\) ft \(/ s\) at an angle \(\alpha\) above the horizontal, then its position after \(t\) seconds is given by the parametric equations $$ x=\left(v_{0} \cos \alpha\right) t \quad y=\left(v_{0} \sin \alpha\right) t-16 t^{2} $$ (where \(x\) and \(y\) are measured in feet). Show that the path of the projectile is a parabola by eliminating the parameter \(t\)

Step-by-Step Solution

Answer

The path of the projectile is a parabola, given by \( y = x \tan \alpha - \frac{16 x^2}{(v_{0} \cos \alpha)^2} \).

Step 1: Write Down the Given Equations

We have two parametric equations that describe the position of the projectile: \[ x = (v_{0} \cos \alpha) t \] and \[ y = (v_{0} \sin \alpha) t - 16t^2 \]. We aim to eliminate the parameter \(t\) to write \(y\) as a function of \(x\).

Step 2: Express t in terms of x

From the given equation \(x = (v_{0} \cos \alpha) t\), solve for \(t\): \[ t = \frac{x}{v_{0} \cos \alpha} \]. This will allow us to substitute into the equation for \(y\).

Step 3: Substitute t into the y Equation

Substitute \(t = \frac{x}{v_{0} \cos \alpha}\) into the equation for \(y\): \[ y = (v_{0} \sin \alpha) \left(\frac{x}{v_{0} \cos \alpha}\right) - 16\left(\frac{x}{v_{0} \cos \alpha}\right)^2 \].

Step 4: Simplify the Equation

Start by simplifying the term \((v_{0} \sin \alpha) \left(\frac{x}{v_{0} \cos \alpha}\right)\): \[ y = \frac{x \tan \alpha}{1} - 16\left(\frac{x^2}{(v_{0} \cos \alpha)^2}\right) \]. This becomes: \[ y = x \tan \alpha - \frac{16 x^2}{(v_{0} \cos \alpha)^2} \].

Step 5: Identify the Parabolic Form

The equation \( y = x \tan \alpha - \frac{16 x^2}{(v_{0} \cos \alpha)^2}\) is quadratic in terms of \(x\), verifying that the path of the projectile is a parabola. This form \((ax + bx^2)\) confirms the parabolic path.

Key Concepts

Projectile Motion

Projectile motion is the motion of an object thrown or propelled into the air, subject to the acceleration due to gravity. It is a form of motion where an object moves along a curved path under the influence of gravity only. Here are some key elements:

• The path followed by the projectile is called its trajectory.
• In simplistic terms, a projectile is any object that is in flight after being thrown, dropped, or launched.
• The motion of a projectile can be analyzed by splitting it into two components: horizontal and vertical.

The horizontal motion involves constant velocity since no force acts in the horizontal direction (ignoring air resistance).
The vertical motion involves constant acceleration caused by gravity.
By using parametric equations like those given in our exercise, we can find the position at any time. This helps us understand how projectiles move and predict their trajectories.

Parabola

A parabola is a U-shaped curve that can describe various motions, such as the path of a projectile in motion. It is one of the basic shapes in algebra and appears naturally in physics when simulating real-world phenomena.A few important properties include:

• The vertex of the parabola is the highest or lowest point of the curve, which can represent the peak height of a projectile.
• Parabolas are always symmetric around a vertical line through the vertex.
• When a projectile is thrown, it follows a parabolic path because the angle and initial speed determine the trajectory shape.

In our context, eliminating the parameter \( t \) from the parametric equations helps us represent the projectile's path as a standard quadratic equation, confirming this shape.
The simplicity of this shape allows for straightforward calculations and predictions of motion.

Quadratic Equations

Quadratic equations are mathematical expressions of the second degree, generally presented in the form \( ax^2 + bx + c = 0 \). The path of a projectile can often be described using such an equation, relating it to the symmetrically curved parabola.Key features of quadratic equations include:

• Their graph is a parabola.
• The coefficients \( a \), \( b \), and \( c \) determine the shape and position of the parabola.
• They can be solved using various methods, such as factoring, the quadratic formula, or completing the square.

In the projectile motion exercise, eliminating time \( t \) led to a quadratic equation in terms of \( x \).
Recognizing the resulting quadratic form tells us that the projectile’s path is parabolic.
This form reveals crucial information like direction, height, and range, making quadratic equations essential tools for analyzing projectile motion.