/*! 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 69 A baseball is thrown with a velo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 69

A baseball is thrown with a velocity of \(31.1 \mathrm{~m} / \mathrm{s}\) at an angle of \(\theta=33.4^{\circ}\) above horizontal. What is the horizontal component of the ball's velocity at the highest point of the ball's trajectory?

Short Answer

Expert verified
Answer: The horizontal component of the ball's velocity at the highest point of its trajectory is approximately 25.96 m/s.

Step by step solution

01

Find the initial velocity components in horizontal and vertical directions

We can find the initial velocity components using the given initial velocity and angle. Initial horizontal component of velocity (u_x): \(u_x = u \cos(\theta)\) Initial vertical component of velocity (u_y): \(u_y = u \sin(\theta)\) where u is the initial velocity (31.1 m/s) and 胃 is the angle (33.4掳).
02

Calculate the initial horizontal component of velocity

Use the formula for the horizontal component: \(u_x = u \cos(\theta)\) \(u_x = 31.1 \times \cos(33.4^{\circ})\) \(u_x \approx 25.96 \mathrm{~m} / \mathrm{s}\) So, the initial horizontal component of the velocity is approximately 25.96 m/s.
03

Calculate the initial vertical component of velocity

Now, use the formula for the vertical component: \(u_y = u \sin(\theta)\) \(u_y = 31.1 \times \sin(33.4^{\circ})\) \(u_y \approx 17.02 \mathrm{~m} / \mathrm{s}\) So, the initial vertical component of the velocity is approximately 17.02 m/s.
04

Identify the horizontal component of the velocity at the highest point of the trajectory

As the horizontal component of the velocity remains constant throughout the entire flight, the horizontal component of the velocity at the highest point of the trajectory will remain the same, i.e., 25.96 m/s. Therefore, the horizontal component of the ball's velocity at the highest point of the trajectory is approximately \(25.96 \mathrm{~m} / \mathrm{s}\).

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 91影视!

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

A rock is thrown at an angle \(45^{\circ}\) below the horizontal from the top of a building. Immediately after release will its acceleration be greater than, equal to, or less than the acceleration due to gravity?

In ideal projectile motion, the velocity and acceleration of the projectile at its maximum height are, respectively, a) horizontal, vertical c) zero, zero. downward. d) zero, vertical downward. b) horizontal, zero. e) zero, horizontal.

To determine the gravitational acceleration at the surface of a newly discovered planet, scientists perform a projectile motion experiment. They launch a small model rocket at an initial speed of \(50.0 \mathrm{~m} / \mathrm{s}\) and an angle of \(30.0^{\circ}\) above the horizontal and measure the (horizontal) range on flat ground to be \(2165 \mathrm{~m}\). Determine the value of \(g\) for the planet.

What is the magnitude of an object's average velocity if an object moves from a point with coordinates \(x=2.0 \mathrm{~m}\) \(y=-3.0 \mathrm{~m}\) to a point with coordinates \(x=5.0 \mathrm{~m}, y=-9.0 \mathrm{~m}\) in a time interval of \(2.4 \mathrm{~s} ?\)

To attain maximum height for the trajectory of a projectile, what angle would you choose between \(0^{\circ}\) and \(90^{\circ}\), assuming that you can launch the projectile with the same initial speed independent of the launch angle. Explain your reasoning.
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Work Energy and Power

Read Explanation

Linear Momentum

Read Explanation

Force

Read Explanation

Torque and Rotational Motion

Read Explanation

Physical Quantities And Units

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.