Question

Project: If a projectile is launched with a horizontal velocity \(v_{x}\) and vertical velocity \(v_{y},\) its position after \(t\) seconds is given by $$\begin{array}{l}x=v_{x} t \\\y=v_{y} t-(g / 2) t^{2}\end{array}$$ where \(g\) is the acceleration due to gravity. Eliminate \(t\) from this pair of equations to get \(y=f(x),\) and show that this equation represents a parabola.

Short answer

By expressing \(t\) as \(t = x / v_{x}\) from the first equation and substituting into the second, and then simplifying, we get \(y = (v_{y} / v_{x}) x - (g / (2 v_{x}^2)) x^2\), which shows that \(y\) as a function of \(x\) is a quadratic equation representing a parabola.

Step-by-step solution

Express t from the x equation

From the first equation for the position of the projectile in the x-direction, solve for the time variable t: Given that \(x = v_{x} t\), we can express \(t\) as \(t = x / v_{x}\).

Substitute t in the y equation

Replace the expression for \(t\) from the first equation into the second equation for y. The second equation is \(y = v_{y} t - (g / 2) t^2\). When we substitute \(t\) we get \(y = v_{y} * (x / v_{x}) - (g / 2) * (x / v_{x})^2\).

Simplify the equation

Simplify the y equation by distributing through and combining like terms. This results in \(y = (v_{y} / v_{x}) x - (g / 2) (x^2 / v_{x}^2)\). This can be further simplified to \(y = (v_{y} / v_{x}) x - (g / (2 v_{x}^2)) x^2\).

Identify the parabolic form

The simplified equation of y we have now is in the form of a parabola, which is \(y = ax^2 + bx + c\). Here, \(a = -(g / (2 v_{x}^2))\), \(b = (v_{y} / v_{x})\), and \(c = 0\). Since a is negative, this particular parabola opens downward.

Key concepts

Parabolic Trajectory

When an object is launched into the air and influenced only by gravity and initial velocities, it follows a parabolic trajectory. This is a curved path that is symmetrical and resembles the shape of a parabola. The motion of projectiles, such as a basketball thrown into a hoop or a stone thrown into a river, typically follows this parabolic trajectory.

In the exercise, we can determine the parabolic trajectory by examining the given equations for horizontal and vertical position over time. The horizontal distance x traveled by a projectile is directly proportional to the horizontal velocity (v_{x}) and the time t it has been in motion. However, the vertical distance y is affected not just by the initial vertical velocity (v_{y}), but also by the constant downward pull of gravity, causing the project to decelerate as it rises and accelerates as it falls back down. This influence of gravity introduces a quadratic term in t to the vertical position equation, resulting in the parabolic shape when graphed.

Understanding this helps students visualize the path of a projectile and relates the abstract equations to the physical world, making it easier to comprehend the motion dynamics of the projectile.

Kinematic Equations

Kinematic equations describe the motion of objects without considering the forces that cause this motion. These equations are used extensively in physics to analyze the motion of projectiles. In our problem, kinematic equations are applied to give us the position of the projectile in both the x (horizontal) and y (vertical) dimensions over time.

The initial velocities in the horizontal (v_{x}) and vertical (v_{y}) directions, along with the downward acceleration due to gravity (g), provide all the necessary information to predict the projectile's path. The equation (x = v_{x} t) highlights that the horizontal motion is uniform, meaning the projectile moves the same distance in the x-direction for every unit of time since there is no horizontal acceleration. On the other hand, the equation (y = v_{y} t - (g / 2) t^2) reveals the changing velocity in the y-direction due to the effect of gravity. By manipulating these kinematic equations, we can better understand the relationship between position, velocity, acceleration, and time.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation of the form (ax^2 + bx + c = 0), where a, b, and c are constants, and a ≠ 0 ensures that the equation is indeed quadratic. The graph of a quadratic equation is a parabola, which can open upwards or downwards depending on the sign of the coefficient a.

In the context of projectile motion, the vertical position equation becomes a quadratic equation when time t is eliminated. This allows us to see that the projectile's vertical motion can be modeled as a parabola with respect to horizontal displacement. The coefficients of this quadratic equation, which we labeled as a, b, and c, correspond to specific physical quantities: a is related to the gravitational acceleration, b to the initial velocity ratio, and c represents initial height which in our problem is zero. Emphasizing this connection helps students learn how standard mathematical concepts, such as the quadratic equation, apply to physical phenomena like projectile motion.

Elimination of Variables

In algebra and physics, elimination of variables is a method we use to reduce the number of variables in an equation or a system of equations. This technique allows us to express a complex relationship in simpler terms, often to solve for a particular variable or understand a concept more deeply.

In our projectile motion problem, eliminating the variable t simplifies the understanding of how the horizontal distance x affects the vertical position y, independently of time. By substituting the expression for t obtained from the horizontal motion equation into the vertical motion equation, we were able to eliminate the variable t and express y solely as a function of x. This simplification is crucial for analyzing the projectile's path without concerning ourselves with the particular time at which the projectile reaches a certain position. This concept is not only essential for solving the problem at hand but also a valuable tool for students solving systems of equations in general.