/*! 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 33 A projectile is launched with an... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 33

A projectile is launched with an initial speed of \(60.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) above the horizontal. The projectile lands on a hillside \(4.00\) s later. Neglect air friction. (a) What is the projectile's velocity at the highest point of its trajectory? (b) What is the straight-line distance from where the projectile was launched to where it hits its target?

Short Answer

Expert verified
The velocity at the highest point of the trajectory is \(51.96 \, m/s\) and the straight-line distance from where the projectile was launched to where it hits its target is \(207.84 \, m\).

Step by step solution

01

Find the initial velocity components

The first thing to do is to decompose the initial speed into horizontal and vertical components using trigonometry. The initial horizontal velocity \(v_{i,x}\) is given by \(v_i \times \cos(\Theta)\), and the initial vertical velocity \(v_{i,y}\) is given by \(v_i \times \sin(\Theta)\). Here, \(v_i = 60.0 \, m/s\) and \(\Theta = 30^{\circ}\). So, \(v_{i,x} = 60 \cos(30^{\circ}) = 51.96 \, m/s\), and \(v_{i,y} = 60 \sin(30^{\circ}) = 30 \, m/s\).
02

Calculate the velocity at the highest point

Now, we will solve part (a) of the problem. At the highest point in its trajectory, the projectile is only moving horizontally. This means the vertical component of the velocity is zero, but the horizontal component is unchanged. Thus, the velocity at the highest point of the trajectory is simply the initial horizontal velocity \(v_{i,x} = 51.96 \, m/s\).
03

Calculate the straight-line distance

To solve part (b), we need to find the range or horizontal distance that the projectile covers. This is simply the horizontal speed multiplied by the time, i.e., Range = \(v_{i,x} \times t\). Where \(v_{i,x} = 51.96 \, m/s\) and \(t = 4.00 \, s\). So, Range = \(51.96 \, m/s \times 4.00 \, s = 207.84 \, m\).

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.

Initial Velocity Decomposition
When dealing with projectile motion, one of the first steps is to break down the initial velocity into horizontal and vertical components. This helps us understand how the projectile moves in each direction. We use basic trigonometry to do this.

For a projectile launched with an initial velocity \(v_i\) at an angle \(\Theta\), the horizontal component \(v_{i,x}\) can be found using \(v_{i,x} = v_i \times \cos(\Theta)\). The vertical component \(v_{i,y}\) is given by \(v_{i,y} = v_i \times \sin(\Theta)\).

In our example, the initial speed is \(60.0 \, m/s\) and the launch angle is \(30^{\circ}\). By applying the formulas:
  • Horizontal Velocity: \(v_{i,x} = 60 \cos(30^{\circ}) = 51.96 \, m/s\)
  • Vertical Velocity: \(v_{i,y} = 60 \sin(30^{\circ}) = 30 \, m/s\)
Breaking down the velocity this way allows us to separately analyze the horizontal and vertical motions of the projectile.
Trajectory Analysis
Understanding the trajectory of a projectile is crucial to find its highest point and other features. The trajectory is the path followed by a moving object. In projectile motion, it's usually a parabolic path.

At the highest point of the projectile's trajectory, the vertical velocity component becomes zero. This is because gravity slows the vertical speed until it momentarily stops at the top. However, the horizontal component remains constant as there is no air resistance to slow it down.

For our problem, at the highest point:
  • Vertical Velocity: 0 \, m/s
  • Horizontal Velocity: \(51.96 \, m/s\)
This helps us determine that the velocity at the highest point is purely horizontal, matching the initial horizontal component.
Horizontal Range Calculation
The horizontal range of a projectile is the distance it travels horizontally between launch and landing. To find this, we multiply the horizontal velocity by the total time the projectile is in the air.

The horizontal velocity \(v_{i,x}\) remains constant during the flight if air friction is neglected. So, the formula for range is:
  • Range = \(v_{i,x} \times t\)

In our case, with \(v_{i,x} = 51.96 \, m/s\) and flight time \(t = 4.00 \, s\), the range becomes:
  • Range: \(51.96 \, m/s \times 4.00 \, s = 207.84 \, m\)
This calculation shows how far the projectile travels horizontally from its starting point.

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 Coast Guard cutter detects an unidentified ship at a distance of \(20.0 \mathrm{~km}\) in the direction \(15.0^{\circ}\) east of north. The ship is traveling at \(26.0 \mathrm{~km} / \mathrm{h}\) on a course at \(40.0^{\circ}\) east of north. The Coast Guard wishes to send a speedboat to intercept and investigate the vessel. (a) If the speedboat travels at \(50.0 \mathrm{~km} / \mathrm{h}\), in what direction should it head? Express the direction as a compass bearing with respect to due north. (b) Find the time required for the cutter to intercept the ship.

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?

An artillery shell is fired with an initial velocity of \(300 \mathrm{~m} / \mathrm{s}\) at \(55.0^{\circ}\) above the horizontal. To clear an avalanche, it explodes on a mountainside \(42.0 \mathrm{~s}\) after firing. What are the \(x\) - and \(y\)-coordinates of the shell where it explodes, relative to its firing point?

S In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strikes the floor at distance \(d\) from the base of the counter. If the height of the counter is \(h\), (a) find an expression for the time \(t\) it takes the cup to fall to the floor in terms of the variables \(h\) and \(g\). (b) With what speed does the mug leave the counter? Answer in terms of the variables \(d, g\), and \(h .\) (c) In the same terms, what is the speed of the cup immediately before it hits the floor? (d) In terms of \(h\) and \(d\), what is the direction of the cup's velocity immediately before it hits the floor?

The record distance in the sport of throwing cowpats is \(81.1 \mathrm{~m}\). This record toss was set by Steve Urner of the United States in 1981. Assuming the initial launch angle was \(45^{\circ}\) and neglecting air resistance, determine (a) the initial speed of the projectile and (b) the total time the projectile was in flight. (c) Qualitatively, how would the answers change if the launch angle were greater than \(45^{\text {" }}\) Explain.
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Solid State Physics

Read Explanation

Turning Points in Physics

Read Explanation

Energy Physics

Read Explanation

Quantum Physics

Read Explanation

Electricity and Magnetism

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.