Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 8

Two cannonballs are shot from different cannons at angles \(\theta_{01}=20^{\circ}\) and \(\theta_{02}=30^{\circ}\), respectively. Assuming ideal projectile motion, the ratio of the launching speeds, \(v_{01} / v_{02},\) for which the two cannonballs achieve the same range is a) \(0.742 \mathrm{~m}\) d) \(1.093 \mathrm{~m}\) b) \(0.862 \mathrm{~m}\) e) \(2.222 \mathrm{~m}\) c) \(1.212 \mathrm{~m}\)

Short Answer

Expert verified
Answer: (b) 0.862

Step by step solution

01

Write down the range of projectile formula

In general, the range R of an object in projectile motion can be represented by the following equation: \(R = \frac{v^2 \sin 2\theta}{g}\) where \(v\) is the initial velocity, \(\theta\) is the launch angle, and \(g\) is the acceleration due to gravity.
02

Define ranges for the two cannonballs

Let's represent the range of the first cannonball \(R_1\) and the range of the second cannonball \(R_2\). According to the problem, \(R_1 = R_2\).
03

Write down the ranges for the two cannonballs

Using the formula from step 1, let's write down the expressions for the ranges of the two cannonballs: \(R_1 = \frac{v_{01}^2 \sin 2\theta_{01}}{g}\) \(R_2 = \frac{v_{02}^2 \sin 2\theta_{02}}{g}\) Replace the angles with the given angles: \(R_1 = \frac{v_{01}^2 \sin 2 \cdot 20^\circ}{g}\) \(R_2 = \frac{v_{02}^2 \sin 2 \cdot 30^\circ}{g}\)
04

Set the ranges equal and solve for the ratio

Now, since \(R_1 = R_2\), let's set the two ranges equal and solve for the ratio \(\frac{v_{01}}{v_{02}}\): \(\frac{v_{01}^2 \sin 2 \cdot 20^\circ}{g} = \frac{v_{02}^2 \sin 2 \cdot 30^\circ}{g}\) Divide by g on both sides and then, divide both sides of the equation by \(v_{02}^2\) and by \(\sin2\cdot30^\circ\): \(\frac{v_{01}^2}{v_{02}^2} = \frac{\sin2\cdot20^\circ}{\sin2\cdot30^\circ}\) Taking the square root of both sides, we get: \(\frac{v_{01}}{v_{02}} = \sqrt{\frac{\sin2\cdot20^\circ}{\sin2\cdot30^\circ}}\)
05

Calculate the final answer and match with choices

Now, plug in the angles and find the final answer: \(\frac{v_{01}}{v_{02}} = \sqrt{\frac{\sin40^\circ}{\sin60^\circ}} \approx 0.862\) Matching our result with the given choices, the correct answer is (b) \(0.862\).

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

An object moves in the \(x y\) -plane. The \(x\) - and \(y\) -coordinates of the object as a function of time are given by the following equations: \(x(t)=4.9 t^{2}+2 t+1\) and \(y(t)=3 t+2 .\) What is the velocity vector of the object as a function of time? What is its acceleration vector at a time \(t=2\) s?

Two swimmers with a soft spot for physics engage in a peculiar race that models a famous optics experiment: the Michelson-Morley experiment. The race takes place in a river \(50.0 \mathrm{~m}\) wide that is flowing at a steady rate of \(3.00 \mathrm{~m} / \mathrm{s} .\) Both swimmers start at the same point on one bank and swim at the same speed of \(5.00 \mathrm{~m} / \mathrm{s}\) with respect to the stream. One of the swimmers swims directly across the river to the closest point on the opposite bank and then turns around and swims back to the starting point. The other swimmer swims along the river bank, first upstream a distance exactly equal to the width of the river and then downstream back to the starting point. Who gets back to the starting point first?

You serve a tennis ball from a height of \(1.8 \mathrm{~m}\) above the ground. The ball leaves your racket with a speed of \(18.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(7.00^{\circ}\) above the horizontal. The horizontal distance from the court's baseline to the net is \(11.83 \mathrm{~m},\) and the net is \(1.07 \mathrm{~m}\) high. Neglect spin imparted on the ball as well as air resistance effects. Does the ball clear the net? If yes, by how much? If not, by how much did it miss?

You passed the salt and pepper shakers to your friend at the other end of a table of height \(0.85 \mathrm{~m}\) by sliding them across the table. They both missed your friend and slid off the table, with velocities of \(5 \mathrm{~m} / \mathrm{s}\) and \(2.5 \mathrm{~m} / \mathrm{s}\), respectively. a) Compare the times it takes the shakers to hit the floor. b) Compare the distance that each shaker travels from the edge of the table to the point it hits the floor.

In a projectile motion, the horizontal range and the maximum height attained by the projectile are equal. a) What is the launch angle? b) If everything else stays the same, how should the launch angle, \(\theta_{0},\) of a projectile be changed for the range of the projectile to be halved?
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Quantum Physics

Read Explanation

Work Energy and Power

Read Explanation

Fluids

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Engineering Physics

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