Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 99

You want to make a plot of the trajectory of a projectile. That is, you want to make a plot of the height \(y\) of the projectile as a function of horizontal distance \(x\). The projectile is launched from the origin with an initial speed \(v_{\mathrm{i}}\) at an angle \(\theta\) above the horizontal. Show that the equation of the trajectory followed by the projectile is $$y=\left(\frac{v_{\text {iy }}}{v_{\text {ix }}}\right) x+\left(\frac{-g}{2 v_{\text {ix }}^{2}}\right) x^{2}$$

Short Answer

Expert verified
Question: Derive the trajectory equation for a projectile launched at an angle θ above the horizontal with an initial speed of 𝑣𝑖. Answer: The trajectory equation for a projectile launched at an angle θ above the horizontal with an initial speed of 𝑣𝑖 is given by \(y=\left(\frac{v_{\text{iy}}}{v_{\text{ix}}}\right) x+\left(\frac{-g}{2v_{\text{ix }}^{2}}\right) x^{2}\), where \(v_{\text{ix}} = v_\mathrm{i} \cos \theta\), and \(v_{\text{iy}} = v_\mathrm{i} \sin \theta\).

Step by step solution

01

Find the horizontal and vertical components of the initial velocity

To find the horizontal and vertical components of the initial velocity, we use the trigonometric functions, sine and cosine, multiplied with the initial speed. These components are given by: $$ v_{\text{ix}} = v_\mathrm{i} \cos \theta \\ v_{\text{iy}} = v_\mathrm{i} \sin \theta $$
02

Write the kinematic equations for horizontal and vertical motion

For the horizontal motion, there is no acceleration, and the equation is given by: $$ x = v_{\text{ix}} t $$ For the vertical motion, the acceleration is the gravitational acceleration \(g\), and we have the following equation: $$ y = v_{\text{iy}}t - \frac{1}{2}gt^2 $$
03

Substitute the time from the horizontal motion equation into the vertical motion equation

Rearrange the horizontal motion equation to isolate \(t\): $$ t=\frac{x}{v_{\text{ix}}} $$ Now, substitute the expression for \(t\) into the vertical motion equation: $$ y = v_{\text{iy}} \frac{x}{v_{\text{ix}}} - \frac{1}{2}g \left(\frac{x}{v_{\text{ix}}}\right)^2 $$
04

Simplify and rewrite the trajectory equation

Simplify the expression to get the desired trajectory equation: $$ y=\left(\frac{v_{\text {iy }}}{v_{\text {ix }}}\right) x+\left(\frac{-g}{2 v_{\text {ix }}^{2}}\right) x^{2} $$ This is the trajectory equation for a projectile launched at an angle \(\theta\) above the horizontal with initial speed \(v_{\mathrm{i}}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Two displacement vectors each have magnitude \(20 \mathrm{km}\) One is directed \(60^{\circ}\) above the \(+x\) -axis; the other is directed \(60^{\circ}\) below the \(+x\) -axis. What is the vector sum of these two displacements? Use graph paper to find your answer.
Jerry bicycles from his dorm to the local fitness center: 3.00 miles east and 2.00 miles north. Cindy's apartment is located 1.50 miles west of Jerry's dorm. If Cindy is able to meet Jerry at the fitness center by bicycling in a straight line, what is the length and direction she must travel?
A sailboat sails from Marblehead Harbor directly east for 45 nautical miles, then \(60^{\circ}\) south of east for 20.0 nautical miles, returns to an easterly heading for 30.0 nautical miles, and sails \(30^{\circ}\) east of north for 10.0 nautical miles, then west for 62 nautical miles. At that time the boat becomes becalmed and the auxiliary engine fails to start. The crew decides to notify the Coast Guard of their position. Using graph paper, ruler, and protractor, sketch a graphical addition of the displacement vectors and estimate their position.
A pilot wants to fly from Dallas to Oklahoma City, a distance of $330 \mathrm{km}\( at an angle of \)10.0^{\circ}$ west of north. The pilot heads directly toward Oklahoma City with an air speed of $200 \mathrm{km} / \mathrm{h}\(. After flying for \)1.0 \mathrm{h},\( the pilot finds that he is \)15 \mathrm{km}$ off course to the west of where he expected to be after one hour assuming there was no wind. (a) What is the velocity and direction of the wind? (b) In what direction should the pilot have headed his plane to fly directly to Oklahoma City without being blown off course?
You will be hiking to a lake with some of your friends by following the trails indicated on a map at the trailhead. The map says that you will travel 1.6 mi directly north, then \(2.2 \mathrm{mi}\) in a direction \(35^{\circ}\) east of north, then finally \(1.1 \mathrm{mi}\) in a direction \(15^{\circ}\) north of east. At the end of this hike, how far will you be from where you started, and what direction will you be from your starting point?
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Dynamics

Read Explanation

Classical Mechanics

Read Explanation

Modern Physics

Read Explanation

Fields in Physics

Read Explanation

Famous Physicists

Read Explanation
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.

Sign-up for free