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Prove that the trajectory of a projectile is parabolic, having the form \(y=a x+b x^{2}\). To obtain this expression, solve the equation \(x=v_{0 x} t\) for \(t\) and substitute it into the expression for \(y=v_{0 y} t-(1 / 2) g t^{2}\) (These equations describe the \(x\) and \(y\) positions of a projectile that starts at the origin.) You should obtain an equation of the form \(y=a x+b x^{2}\) where \(a\) and \(b\) are constants.

Short Answer

Expert verified
The trajectory of a projectile is parabolic when time \(t\) from the horizontal displacement equation \(x = v_{0x} t\) is substituted into the vertical displacement equation \(y = v_{0y} t - \frac{1}{2} g t^2\), simplifying to the form \(y = ax + bx^2\) with \(a = \frac{v_{0y}}{v_{0x}}\) and \(b = -\frac{g}{2v_{0x}^2}\).

Step by step solution

01

Identify the Given Equations

We are given the horizontal motion equation for a projectile, which is: \(x = v_{0x} t\), where \(v_{0x}\) is the initial horizontal velocity, and \(t\) is the time elapsed. We are also given the vertical motion equation: \(y = v_{0y} t - \frac{1}{2}gt^2\), where \(v_{0y}\) is the initial vertical velocity, and \(g\) is the acceleration due to gravity (9.8 m/s^2 and is considered to be constant).
02

Solve for Time \(t\)

From the horizontal motion equation \(x = v_{0x} t\), we can solve for time \(t\) by dividing both sides of the equation by \(v_{0x}\): \(t = \frac{x}{v_{0x}}\).
03

Substitute Time \(t\) into the Vertical Motion Equation

Substitute the expression for \(t\) into the vertical motion equation: \[y = v_{0y} \left(\frac{x}{v_{0x}}\right) - \frac{1}{2}g \left(\frac{x}{v_{0x}}\right)^2\]. By doing this substitution, we will express \(y\) as a function of \(x\) only.
04

Simplify the Equation

Simplify the equation by distributing and combining like terms to find constants \(a\) and \(b\): \[y = \frac{v_{0y}}{v_{0x}} x - \frac{g}{2v_{0x}^2} x^2\]. Here, \(\frac{v_{0y}}{v_{0x}}\) becomes the constant \(a\) and \(-\frac{g}{2v_{0x}^2}\) becomes the constant \(b\), resulting in the parabolic form \(y = ax + bx^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parabolic Trajectory
The concept of parabolic trajectory is fascinating and a fundamental part of classical mechanics. When we talk about a projectile, we often imagine an object being thrown into the air and following a certain path before landing. This path is called its trajectory, and under the influence of gravity alone, without any air resistance, this path is parabolic in shape.

But what makes the trajectory parabolic? It’s the constant acceleration due to gravity acting on the projectile. Despite gravity pulling the projectile downwards, it initially has some vertical and horizontal velocities, causing it to move both forward and upward. This motion continues until gravity overcomes the object's initial upward momentum, forcing it downward into a descending arc.

As a point mass—a simplification where we consider the object as if all its mass is concentrated at a single point—is projected forward, the only acceleration it experiences is the gravitational pull towards the Earth. This constant acceleration creates a symmetric path, and when plotted on a graph where the horizontal axis (x) represents distance and the vertical axis (y) represents height, the resulting curve is a parabola.

In the exercise you’re dealing with, it’s shown mathematically that the projectile's position defined by the equations for horizontal and vertical motions leads to a quadratic equation in terms of x, confirming the shape is indeed parabolic. This understanding not only helps in academic learning but also has real-world applications in fields like ballistics, sports, and even spaceflight.
Kinematic Equations
Understanding kinematic equations is vital for solving any physics problem that involves motion. These equations relate the five kinematic variables: displacement, initial velocity, final velocity, acceleration, and time. In projectile motion, we primarily use these equations to find the position of the projectile at any point in time.

To delve deeper, let’s break down the two dimensions of motion. Horizontally, the projectile doesn’t accelerate (ignoring air resistance), so its horizontal velocity remains constant. Vertically, it experiences acceleration due to gravity. The kinematic equations you’ve used in the exercise – often referred to as the 'equations of motion' – are tailored specifically for these conditions.

By isolating time from the horizontal motion equation and plugging it into the vertical motion equation, you switch to describing vertical position (y) as a function of horizontal position (x) – this is an important strategy in physics problems. It aligns perfectly with the definition of a function in mathematics, where every value of x corresponds to exactly one value of y, drawing out that familiar parabolic trajectory.

When you master these fundamental equations, you can apply them to a vast array of problems in physics, from the simplest to the most complex. They serve as building blocks for understanding motion, and grasping their use is crucial for success in the subject.
Physics Problem Solving
When it comes to physics problem solving, a step-by-step approach is often the most effective method. Starting with a clear understanding of the problem, isolating known and unknown variables, and then systematically applying physical laws and equations will help facilitate a solution.

In the projectile motion exercise, the problem-solving method is particularly clear. After identifying the kinematic equations for horizontal and vertical motion, the next step is to solve for a variable that connects these two motions—in this case, time (t). Isolating this variable allows you to eliminate it from the equations, thereby finding a relationship solely between the vertical and horizontal positions.

This approach not only simplifies the problem but also shapes your thinking towards more complex concepts like eliminating parameters to find a direct relationship between variables. It’s through this methodology that physics transcends the classroom, allowing us to formulate predictions and understand the world around us. Moreover, by analyzing the trajectory of a projectile, students can relate to real-life motions they observe, making physics less abstract and more tangible. Whether it’s throwing a ball or designing a roller coaster, the principles remain the same and are grounded in the step-by-step methods of problem-solving in physics.

Encouraging students to take these steps—conceptual understanding, equation manipulation, and strategic substitution—enhances their problem-solving skills and boosts confidence in tackling physics challenges.

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Most popular questions from this chapter

Give an example of a nonzero vector that has a component of zero.

Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither \(0^{\circ}\) nor \(90^{\circ}\) ): (a) Is the acceleration ever zero? (b) Is the acceleration ever in the same direction as a component of velocity? (c) Is the acceleration ever opposite in direction to a component of velocity?

Construct Your Own Problem Consider a ball tossed over a fence. Construct a problem in which you calculate the ball's needed initial velocity to just clear the fence. Among the things to determine are; the height of the fence, the distance to the fence from the point of release of the ball, and the height at which the ball is released. You should also consider whether it is possible to choose the initial speed for the ball and just calculate the angle at which it is thrown. Also examine the possibility of multiple solutions given the distances and heights you have chosen.

Explain why it is not possible to add a scalar to a vector.

A ball is thrown horizontally from the top of a \(60.0\) -m building and lands \(100.0 \mathrm{~m}\) from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

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