/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 The equation of a parabola is \(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The equation of a parabola is \(y=a x^{2}+b x+c\), where \(a, b\), and care constants. The \(x\)-and \(y\)-coordinates of a projectile launched from the origin as a function of time are given by \(x=v_{0 x} t\) and \(y=v_{0 y} t-\frac{1}{2} g t^{2}\), where \(v_{0 x}\) and \(v_{0 y}\) are the components of the initial velocity. (a) Eliminate \(t\) from these two equations and show that the path of a projectile is a parabola and has the form \(y=a x+b x^{2}\). (b) What are the values of \(a, b\), and \(c\) for the projectile?

Short Answer

Expert verified
The trajectory of the projectile is a parabola described by \(y= -\frac{gx^{2}}{2v_{0x}^{2}} + \frac{v_{0y}}{v_{0x}}x\), where \(a = -\frac{g}{2v_{0x}^{2}}, b = \frac{v_{0y}}{v_{0x}}\) and \(c = 0\).

Step by step solution

01

Eliminating Time (t) From Equations

We begin with the two equations of motion: \(x = v_{0x}t\) and \(y = v_{0y}t - \frac{1}{2}gt^{2}\). First rearrange the equation for x to isolate t, i.e. \(t=\frac{x}{v_{0x}}\). Then substitute this into the equation for y.
02

Simplifying and Identifying the Parabola Form

Substituting the equation from Step 1 into the equation for y, we get \(y = v_{0y}\(\frac{x}{v_{0x}}\) - \frac{1}{2}g(\(\frac{x}{v_{0x}}\))^{2}\). Simplifying further, we obtain \(y = \frac{v_{0y}}{v_{0x}}x - \frac{gx^{2}}{2v_{0x}^{2}}\), which takes the form \(y = Ax + Bx^{2}\), the equation for a parabola.
03

Identifying the Coefficients ('a', 'b' and 'c') Of the Parabolic Path

The equation from Step 2 is an equation of a parabola. Hence, the values of a, b and c can be identified by comparing it with the standard form \(y = ax^{2} + b'x + c\). Thus, in this context, \(a = -\frac{g}{2v_{0x}^{2}}, b = \frac{v_{0y}}{v_{0x}}\) and \(c = 0\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Parabola Equation
Imagine a smooth arch that you see when you toss a ball up in the air. This arch is described by the equation of a parabola. A parabola is a U-shaped curve on a graph, and its equation is typically written as \(y = ax^2 + bx + c\). Here, \(a\), \(b\), and \(c\) are constants that shape the parabola. The equation shows how the \(y\)-coordinate is related to the \(x\)-coordinate squared, as well as the first power of \(x\).
In projectile motion, this kind of equation helps us understand the path of an object being thrown or projected. By rearranging the motion equations of a projectile to eliminate time \(t\), we get an expression for \(y\) in terms of \(x\). When simplified, this shows the trajectory of the projectile is indeed a parabola. Thus, understanding the parabola equation is crucial in predicting where and how an object will move in projectile motion.
The constants \(a\), \(b\), and \(c\) are found by comparing this trajectory equation with the standard parabolic form.
Kinematics
Kinematics is like the storyteller of physics, describing how things move without diving into why they move that way. In kinematics, we focus on displacement, velocity, and acceleration.
When we talk about projectile motion, kinematics gives us the formulas for the position of a projectile over time. For a projectile, the horizontal position \(x\) is given by \(x = v_{0x}t\), meaning it's simply the horizontal velocity multiplied by time. If there's no horizontal acceleration, this movement is uniform.
For vertical motion, things are a bit different because gravity comes into play. The vertical position \(y\) is described by \(y = v_{0y}t - \frac{1}{2}gt^2\), accounting for both the initial velocity and the acceleration due to gravity. These expressions in kinematics help us predict the path that any thrown or projected object will follow.
Equations of Motion
The equations of motion are powerful tools in physics, allowing us to calculate different aspects of an object's movement. These are especially handy in analyzing projectile motion.
In projectile motion, the main equations are:
  • Horizontal motion: \(x = v_{0x}t\)
  • Vertical motion: \(y = v_{0y}t - \frac{1}{2}gt^2\)
These equations reveal how the object's location changes over time. The combined effect of both horizontal and vertical components gives the object its parabolic path. By eliminating the time variable \(t\) from these equations, we derive a direct relationship between \(x\) and \(y\), showcasing that the path of the projectile traces a parabola.
Understanding these motion equations helps in identifying patterns in motion and predicting future positions and velocities of the moving objects.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An airplane flies \(200 \mathrm{~km}\) due west from city A to city \(\mathrm{B}\) and then \(300 \mathrm{~km}\) in the direction of \(30.0^{\circ}\) north of west from city \(B\) to city \(C\). (a) In straight-line distance, how far is city C from city A? (b) Relative to city \(\mathrm{A}\), in what direction is city \(\mathrm{C}\) ? (c) Why is the answer only approximately correct?

A truck loaded with cannonball watermelons stops suddenly to avoid running over the edge of a washed-out bridge (Fig. P3.74). The quick stop causes a number of melons to fly off the truck. One melon rolls over the edge with an initial speed \(v_{i}=10.0 \mathrm{~m} / \mathrm{s}\) in the horizontal direction. A cross section of the bank has the shape of the bottom half of a parabola with its vertex at the edge of the road, and with the equation \(y^{2}=(16.0 \mathrm{~m}) x\), where \(x\) and \(y\) are measured in meters. What are the \(x\)-and \(y\)-coordinates of the melon when it splatters on the bank?

A daredevil is shot out of a cannon at \(45.0^{\circ}\) to the horizontal with an initial speed of \(25.0 \mathrm{~m} / \mathrm{s}\). A net is positioned a horizontal distance of \(50.0 \mathrm{~m}\) from the cannon. At what height above the cannon should the net be placed in order to catch the daredevil?

M A place-kicker must kick a football from a point \(36.0 \mathrm{~m}\) (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is \(3.05 \mathrm{~m}\) high. When kicked, the ball leaves the ground with a speed of \(20.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(53.0^{\circ}\) to the horizontal. (a) By how much does the ball clear or fall short of clearing the crossbar? (b) Does the ball approach the crossbar while still rising or while falling?

A student decides to measure the muzzle velocity of a pellet shot from his gun. He points the gun horizontally. He places a target on a vertical wall a distance \(x\) away from the gun. The pellet hits the target a vertical distance \(y\) below the gun. (a) Show that the position of the pellet when traveling through the air is given by \(y=A x^{2}\), where \(A\) is a constant. (b) Express the constant \(A\) in terms of the initial (muzzle) velocity and the freefall acceleration. (c) If \(x=3.00 \mathrm{~m}\) and \(y=0.210 \mathrm{~m}\), what is the initial speed of the pellet?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.