/*! 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 15 A skateboarder shoots off a ramp... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 15

A skateboarder shoots off a ramp with a velocity of \(6.6 \mathrm{m} / \mathrm{s},\) directed at an angle of \(58^{\circ}\) above the horizontal. The end of the ramp is \(1.2 \mathrm{m}\) above the ground. Let the \(x\) axis be parallel to the ground, the \(+y\) direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

Short Answer

Expert verified
(a) 5.0 m; (b) 2.0 m from the ramp.

Step by step solution

01

Identify Initial Conditions

The skateboarder has an initial velocity of \( v_0 = 6.6 \, \text{m/s} \). The ramp is inclined at an angle of \( \theta = 58^\circ \). The initial height above the ground \( y_0 = 1.2 \, \text{m} \). We need to break down the velocity into horizontal and vertical components.
02

Resolve Velocity into Components

The horizontal velocity component \( v_{0x} \) is given by \( v_{0x} = v_0 \cos\theta \). Calculate it: \( v_{0x} = 6.6 \cos(58^\circ) \approx 3.5 \, \text{m/s} \). The vertical velocity component \( v_{0y} \) is given by \( v_{0y} = v_0 \sin\theta \). Calculate it: \( v_{0y} = 6.6 \sin(58^\circ) \approx 5.6 \, \text{m/s} \).
03

Calculate Maximum Height Above the Ramp

To find the maximum height, use the vertical motion equation without time: \( v_y^2 = v_{0y}^2 - 2g(y - y_0) \), where \( v_y = 0 \) at the highest point. Solve for \( y \): \( 0 = (5.6)^2 - 2(9.8)(y - 1.2) \), which gives \( y \approx 3.8 \, \text{m} \).
04

Add Initial Height to Get Total Maximum Height

Since the ramp is already 1.2 meters above the ground, add the height above the ramp to find the total maximum height: \( y_{total} = 3.8 + 1.2 = 5.0 \, \text{m} \).
05

Calculate Horizontal Distance at Maximum Height

Use the horizontal motion equation \( x = v_{0x} \times t \). Find the time \( t \) it takes to reach maximum height using vertical motion: \( v_y = v_{0y} - gt \), with \( v_y = 0 \) at the top. Thus, \( 0 = 5.6 - 9.8t \), giving \( t \approx 0.57 \, \text{s} \). Calculate \( x \): \( x = 3.5 \times 0.57 \approx 2.0 \, \text{m} \).

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
In projectile motion, the initial velocity is extremely important because it helps determine the path and distance the object will travel. For the skateboarder, the initial velocity is given as \( 6.6 \, \text{m/s} \) at an angle of \( 58^\circ \) above the horizontal. To thoroughly understand the effect of initial velocity, it's crucial to break it into two components:
  • **Horizontal Component**: This component, \( v_{0x} \), is found using \( v_0 \cos \theta \). It influences how far the object will travel horizontally. For our skateboarder, it is approximately \( 3.5 \, \text{m/s} \).
  • **Vertical Component**: This component, \( v_{0y} \), is calculated with \( v_0 \sin \theta \). It determines how high the object will rise before gravity pulls it back down. Here, the vertical component is \( 5.6 \, \text{m/s} \).
The initial velocity and its components help define the entire motion of the skateboarder as they travel through the air.
Maximum Height
The maximum height in projectile motion is the highest vertical position reached by the object in the air, and it occurs when the vertical velocity component becomes zero. To calculate this, we utilize the principles of kinematics, focusing solely on the vertical motion.For the skateboarder's case:
  • We use the equation \( v_y^2 = v_{0y}^2 - 2g(y - y_0) \), where \( v_y = 0 \) at the highest point. Solving it provides us \( y \approx 3.8 \, \text{m} \) above the ramp.
  • The ramp's top is already \( 1.2 \, \text{m} \) above ground level. Thus, adding this gives the total maximum height from the ground: \( y_{total} = 3.8 + 1.2 = 5.0 \, \text{m} \).
Understanding the maximum height helps in analyzing how high the skateboarder will go, which is particularly important in evaluating obstacles or potential energy considerations.
Horizontal Distance
In the context of projectile motion, the horizontal distance represents how far an object travels along the horizontal plane. It depends mainly on the horizontal component of the initial velocity and the time the object spends in the air.To calculate the distance the skateboarder covers when reaching the highest point:
  • First, determine the time to reach that point by setting the vertical velocity at the highest point to zero using \( v_y = v_{0y} - gt \). Solving this gives \( t \approx 0.57 \, \text{s} \).
  • The horizontal distance \( x \) traversed can be found with \( x = v_{0x} \times t \). For the skateboarder, this results in \( x \approx 2.0 \, \text{m} \).
This distance informs us about the skateboarder's reach and helps predict the landing zone, ensuring safety and hazard avoidance in practical applications.

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

Useful background for this problem can be found in Multiple-Concept Example 2. On a spacecraft two engines fire for a time of 565 s. One gives the craft an acceleration in the \(x\) direction of \(a_{x}=5.10 \mathrm{m} / \mathrm{s}^{2},\) while the other produces an acceleration in the \(y\) direction of \(a_{y}=7.30 \mathrm{m} / \mathrm{s}^{2} .\) At the end of the firing period, the craft has velocity components of \(v_{x}=3775 \mathrm{m} / \mathrm{s}\) and \(v_{y}=4816 \mathrm{m} / \mathrm{s} .\) Find the magnitude and direction of the initial velocity. Express the direction as an angle with respect to the \(+x\) axis.

In a marathon race Chad is out in front, running due north at a speed of \(4.00 \mathrm{m} / \mathrm{s} .\) John is \(95 \mathrm{m}\) behind him, running due north at a speed of 4.50 m/s. How long does it take for John to pass Chad?

A major-league pitcher can throw a baseball in excess of \(41.0 \mathrm{m} / \mathrm{s}\). If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is \(17.0 \mathrm{m}\) away from the point of release?

A projectile is launched from and returns to ground level, as the figure shows. There is no air resistance. The horizontal range of the projectile is measured to be \(R=175 \mathrm{m},\) and the horizontal component of the launch velocity is \(v_{0 x}=+25 \mathrm{m} / \mathrm{s} .\) Concepts: (i) What is the final value of the horizontal component \(v_{x}\) of the projectile's velocity? (ii) Can the time be determined for the horizontal part of the motion? (iii) Is the time for the horizontal part of the motion the same as that for the vertical part? (iv) For the vertical part of the motion, what is the displacement of the projectile? Calculations: Find the vertical component \(v_{0 y}\) of the projectile.

A space vehicle is coasting at a constant velocity of \(21.0 \mathrm{m} / \mathrm{s}\) in the \(+y\) direction relative to a space station. The pilot of the vehicle fires a \(\mathrm{RCS}\) (reaction control system) thruster, which causes it to accelerate at \(0.320 \mathrm{m} / \mathrm{s}^{2}\) in the \(+x\) direction. After \(45.0 \mathrm{s},\) the pilot shuts off the \(\mathrm{RCS}\) thruster. After the RCS thruster is turned off, find (a) the magnitude and (b) the direction of the vehicle's velocity relative to the space station. Express the direction as an angle measured from the \(+y\) direction.
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Circular Motion and Gravitation

Read Explanation

Astrophysics

Read Explanation

Magnetism

Read Explanation

Solid State Physics

Read Explanation

Electromagnetism

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.