/*! 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 23 A projectile is fired horizontal... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 23

A projectile is fired horizontally from a gun that is \(45.0 \mathrm{~m}\) above flat ground, emerging from the gun with a speed of \(250 \mathrm{~m} / \mathrm{s} .\) (a) How long does the projectile remain in the air? (b) At what horizontal distance from the firing point does it strike the ground? (c) What is the magnitude of the vertical component of its velocity as it strikes the ground?

Short Answer

Expert verified
(a) 3.03 s, (b) 757.5 m, (c) 29.7 m/s

Step by step solution

01

Determine the time in the air

To find how long the projectile remains in the air, we use the vertical motion equation for free-fall: \[ h = \frac{1}{2} g t^2 \]where \( h = 45.0 \text{ m} \) is the height, and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. Solve for \( t \):\[ 45.0 = \frac{1}{2} \times 9.8 \times t^2 \]\[ 45.0 = 4.9 t^2 \]\[ t^2 = \frac{45.0}{4.9} \approx 9.18 \]\[ t = \sqrt{9.18} \approx 3.03 \text{ seconds} \]
02

Calculate the horizontal distance

The horizontal distance is determined by the horizontal velocity and the time in the air. Use \[ d = v_x \times t \] where \( v_x = 250 \text{ m/s} \) and \( t = 3.03 \text{ seconds} \):\[ d = 250 \times 3.03 \approx 757.5 \text{ m} \]
03

Find the vertical component of velocity upon impact

To find the vertical component of the velocity as the projectile strikes the ground, use the equation for velocity under constant acceleration:\[ v_y = g \times t \] where \( g = 9.8 \text{ m/s}^2 \) and \( t = 3.03 \text{ seconds} \):\[ v_y = 9.8 \times 3.03 \approx 29.7 \text{ m/s} \]

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!

Key Concepts

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

Horizontal Distance
Did you ever wonder how far a projectile can go when fired horizontally? The horizontal distance traveled depends on how fast the projectile is moving horizontally and how long it stays in the air. Consider the formula
  • \( d = v_x \times t \)
where \( d \) is the horizontal distance, \( v_x \) is the horizontal velocity, and \( t \) is the time. For our projectile, fired at a speed of \(250 \, \text{m/s}\), this distance is calculated by multiplying by the time it stays airborne, which is \(3.03 \, \text{seconds}\). This results in a total horizontal distance of approximately \(757.5 \, \text{meters}\).

Horizontal velocity remains constant because there is no acceleration affecting it in the horizontal direction, assuming air resistance is negligible. This means you simply need to know how long the projectile is in motion to find out how far it travels horizontally.
Vertical Velocity
Vertical velocity in projectile motion is a bit more dynamic. While horizontal movement is steady, vertical movement changes due to gravity. As a projectile falls, its vertical velocity increases over time. When calculating this, we use the formula
  • \( v_y = g \times t \)
where \( v_y \) is the vertical component of velocity, \( g \) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)), and \( t \) is the time in seconds. In our example, after \(3.03\) seconds, the vertical velocity reaches around \(29.7 \, \text{m/s}\) just before impact.

This increasing velocity occurs because gravity constantly accelerates the projectile downward. Understanding vertical velocity is crucial because it helps predict when the projectile will hit the ground and at what speed!
Free-Fall Equations
The intriguing moment when a projectile begins its descent is all about one powerful force: gravity. When an object is in free-fall, we apply free-fall equations to predict its motion, particularly its fall time. With our exercise, the formula
  • \( h = \frac{1}{2} g t^2 \)
was crucial in discovering the time \(t\) the projectile stays airborne. Here, \(h\) is the height (\(45.0\) meters in our case), and \(g\) represents the unyielding pull of gravity, \(9.8 \, \text{m/s}^2\).

Solving for time \(t\) involves isolating it in the equation: \( t = \sqrt{\frac{2h}{g}} \), resulting in \(3.03\) seconds for our problem. Understanding these free-fall equations can make calculating fall time much easier, ensuring accurate results when studying the projectile's journey. Learning how to apply these equations will serve you well in various physics problems related to motion.

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 baseball is hit at ground level. The ball reaches its maximum height above ground level \(3.0 \mathrm{~s}\) after being hit. Then \(2.5 \mathrm{~s}\) after reaching its maximum height, the ball barely clears a fence that is \(97.5 \mathrm{~m}\) from where it was hit. Assume the ground is level. (a) What maximum height above ground level is reached by the ball? (b) How high is the fence? (c) How far beyond the fence does the ball strike the ground?

Snow is falling vertically at a constant speed of \(8.0 \mathrm{~m} / \mathrm{s}\). At what angle from the vertical do the snowflakes appear to be falling as viewed by the driver of a car traveling on a straight, level road with a speed of \(50 \mathrm{~km} / \mathrm{h} ?\)

An airport terminal has a moving sidewalk to speed passengers through a long corridor. Larry does not use the moving sidewalk; he takes \(150 \mathrm{~s}\) to walk through the corridor. Curly, who simply stands on the moving sidewalk, covers the same distance in 70 s. Moe boards the sidewalk and walks along it. How long does Moe take to move through the corridor? Assume that Larry and Moe walk at the same speed.

A boy whirls a stone in a horizontal circle of radius \(1.5 \mathrm{~m}\) and at height \(2.0 \mathrm{~m}\) above level ground. The string breaks, and the stone flies off horizontally and strikes the ground after traveling a horizontal distance of \(10 \mathrm{~m}\). What is the magnitude of the centripetal acceleration of the stone during the circular motion?

A wooden boxcar is moving along a straight railroad track at speed \(v_{1}\). A sniper fires a bullet (initial speed \(v_{2}\) ) at it from a high-powered rifle. The bullet passes through both lengthwise walls of the car, its entrance and exit holes being exactly opposite each other as viewed from within the car. From what direction, relative to the track, is the bullet fired? Assume that the bullet is not deflected upon entering the car, but that its speed decreases by \(20 \%\). Take \(v_{1}=85 \mathrm{~km} / \mathrm{h}\) and \(v_{2}=650 \mathrm{~m} / \mathrm{s}\). (Why don't you need to know the width of the boxcar?)
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Famous Physicists

Read Explanation

Fundamentals of Physics

Read Explanation

Electromagnetism

Read Explanation

Medical Physics

Read Explanation

Classical Mechanics

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.