/*! 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 63 Leaping the River I. A physics p... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 63

Leaping the River I. A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle (Fig. 3.51\() .\) The takeoff ramp was inclined at \(53.0^{\circ}\) , the river was 40.0 \(\mathrm{m}\) wide, and the far bank was 15.0 \(\mathrm{m}\) lower than the top of the ramp. The river itself was 100 \(\mathrm{m}\) below the ramp. You can ignore air resistance. (a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank? (b) If his speed was only half the value found in (a), where did he land?

Short Answer

Expert verified
(a) 21.6 m/s (b) He lands in the river before reaching the other bank.

Step by step solution

01

Understanding the Problem

We are tasked with determining the minimum speed required for the professor to jump across the river and determining where he would land if his speed was half of that required speed. We need to figure out the horizontal and vertical components and use the projectile motion equations.
02

Determine the Initial Speed Required (a)

First, break down the motion into horizontal and vertical components. We want to find the initial speed, \( v_0 \), so that the motorcycle lands 15 m lower on the opposite bank, 40 m horizontally away. 1. Use the horizontal motion equation: \( x = v_{0x} t \rightarrow 40 = v_{0} \cos 53^{\circ} \times t \).2. Use the vertical motion equation with the final position being 15 m below: \( y = v_{0y} t - \frac{1}{2} g t^2 \rightarrow -15 = v_{0} \sin 53^{\circ} \times t - \frac{1}{2} \times 9.81 \times t^2 \).3. Solve these equations simultaneously to find \( v_0 \).
03

Solving Equations Simultaneously for Speed

Substitute \( t \) from the horizontal component into the vertical motion equation. From \( t = \frac{40}{v_{0} \cos 53^{\circ}} \), substitute into the vertical equation:\[ -15 = v_0 \sin 53^{\circ} \left(\frac{40}{v_0 \cos 53^{\circ}}\right) - \frac{1}{2} \cdot 9.81 \cdot \left(\frac{40}{v_0 \cos 53^{\circ}}\right)^2 \]Solve for \( v_0 \).
04

Finding the Value of Initial Speed

Solving the equation, we use trigonometric identities and values for calculations: \( \sin 53 = 0.7986 \) and \( \cos 53 = 0.6018 \).The equation simplifies to solve for \( v_0 \):\[ v_0 = \sqrt{\frac{40^2\cdot 9.81}{2\cdot (40 \tan 53 - 15) }} \approx \sqrt{\frac{1600 \cdot 9.81}{40\cdot 1.327 - 15}} \approx 21.6 \text{ m/s} \]
05

Determine Landing Position (b)

If the speed is half, \( v_0 = 10.8\, \text{m/s} \).1. Use the same horizontal time equation: \( t = \frac{40}{10.8 \cdot \cos 53^{\circ}} \).2. Calculate the time before the motorcycle lands 100 m below the ramp (into the river): \( y = 0 - 100 = 10.8\sin 53^{\circ} t - 4.905t^2 \).3. Solve for new \( t \) and substitute back to find horizontal landing point.
06

Calculating Landing Position into River

For \( y = -100 \): Solve \( -100 = 10.8 \cdot 0.7986\cdot t - 4.905\cdot t^2 \).Using the quadratic formula to solve for \( t \): \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \text{ where } b = 10.8 \cdot 0.7986, a = -4.905, c = 100 \]The positive \( t \) value gives the time until he hits the river.Calculate \( x \) using \( x = 10.8 \times \cos 53^{\circ} \times t \).

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.

Physics Problem Solving
Physics problem solving is an essential skill in tackling exercises such as the daredevil stunt of leaping across a river on a motorcycle. It involves understanding the underlying principles and applying them logically to reach a solution. To approach a problem systematically, follow these steps:
  • Understand the Problem: Carefully read the problem statement to identify what is given and what needs to be found. For example, the motorcycle stunt required the calculation of initial speed and landing position.
  • Break Down the Motion: In projectile motion, the motion can be divided into horizontal and vertical components. This simplifies the process as each component can be analyzed separately using different equations.
  • Apply Relevant Equations: Use appropriate physics equations for each component. For horizontal motion, the equation is usually distance equals horizontal velocity times time. Vertical motion involves using acceleration due to gravity.
  • Solve Mathematically: After setting up equations, use algebraic manipulation and substitutions to derive the unknowns. Employ mathematical tools like trigonometry and the quadratic equation when necessary.
  • Verify the Solution: Once calculations are complete, check the results against the problem's conditions to ensure accuracy.
Approaching physics problems with a clear process helps in understanding and solving even complex challenges efficiently.
Kinematics
Kinematics is a branch of physics that deals with the description of motion without considering the forces that cause it. In the context of leaping across a river, understanding kinematics is crucial to solving how far and how fast one must move.The primary concepts in kinematics include:
  • Displacement: The change in position of an object. For the motorcycle, it involves determining how far horizontally it must travel to reach the opposite bank.
  • Velocity: The rate of change of displacement. Initial velocity at the ramp plays a key role in reaching the other side.
  • Acceleration: Usually involves gravity in projectile motion, acting downwards at 9.81 m/s虏.Kinematic equations describe these relationships, such as:
    • Horizontal motion: \( x = v_{0x} t \)
    • Vertical motion: \( y = v_{0y} t - \frac{1}{2} g t^2 \)
Kinematics allows the breakdown of complicated motions such as the projectile path of the motorcycle, simplifying analysis and calculation.
Trigonometry in Physics
Trigonometry is an invaluable tool in physics, particularly when dealing with problems involving angled components, such as the motorcycle ramp inclined at a specific angle. This math concept helps resolve vectors into components.Key trigonometric elements in physics include:
  • Angles and Their Functions: The sine and cosine functions are critical in determining the horizontal and vertical components of motion. For the ramp inclined at \(53^{\circ}\), use
    • \( \cos 53^{\circ} \approx 0.6018 \): Helps find horizontal components
    • \( \sin 53^{\circ} \approx 0.7986 \): Helps find vertical components
  • Breaking Down Vectors: Any velocity or force at an angle can be split into perpendicular components using trigonometric identities, enabling easier manipulation and equation solving.
  • Solving Problems: Through substitutions and rearrangements, trigonometry integrates with physics equations to derive unknown quantities like the initial speed needed to make the jump without missing the landing zone.
Understanding trigonometry's application in physics enhances problem-solving capabilities, especially in kinematic and projectile motion scenarios, by offering precise ways to calculate angled movements.

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

Centrifuge on Mercury. A laboratory centrifuge on earth makes \(n\) rpm \((\operatorname{rev} / \min )\) and produces an acceleration of 5.00\(g\) at its outer end. (a) What is the acceleration (in \(g^{\prime}\) s) at a point halfway out to the end? (b) This centrifuge is now used in a space capsule on the planet Mercury, where \(g_{\text { Mexary }}\) is 0.378 what it is on earth. How many \(\operatorname{rom}\) (in terms of \(n )\) should it make to produce 5g at its outer end?

The coordinstes of a bird flying in the \(x y\) -plane are given by \(x(t)=\alpha t\) and \(y(t)=3.0 \mathrm{m}-\beta t^{2},\) where \(\alpha=2.4 \mathrm{m} / \mathrm{s}\) and \(\beta=1.2 \mathrm{m} / \mathrm{s}^{2} \cdot\) (a) Sketch the path of the bird between \(t=0\) and \(t=2.0 \mathrm{s}\) . (b) Calculate the velocity and acceleration vectors of the bird as functions of time. (c) Calculate the magnitude and direction of the bird's velocity and acceleration at \(t=20 \mathrm{s}\) . (d) Sketch the velocity and acceleration vectors at \(t=2.0 \mathrm{s}\) . At this instant, is the bird speeding up, is it slowing down, or is its speed instantaneously not changing? Is the bird turning? If so, in what direction?

The radius of the earth's orbit around the sun (assumed to be circular) is \(1.50 \times 10^{8} \mathrm{km},\) and the earth travels around this orbit in 365 days. (a) What is the magnitude of the orbital velocity of the earth, in \(\mathrm{m} / \mathrm{s} ?\) (b) What is the radial acceleration of the earth toward the sun, in \(\mathrm{m} / \mathrm{s}^{2} ?\) (c) Repeat parts (a) and (b) for the motion of the planet Mercury (orbit radius \(=5.79 \times 10^{7} \mathrm{km}\) , orbital period \(=88.0\) days.

Two crickets, Chirpy and Milada, jump from the top of a vertical cliff. Chirpy just drops and reaches the ground in 3.50 \(\mathrm{s}\) , while Milada jumps horizontally with an initial speed of 95.0 \(\mathrm{cm} / \mathrm{s}\) . How far from the base of the cliff will Milada hit the ground?

Cycloid. A particle moves in the \(x y\) -plane. Its coordinates are given as functions of time by $$ x(t)=R(\omega t-\sin \omega t) \quad y(t)=R(1-\cos \omega t) $$ where \(R\) and \(\omega\) are constants. (a) Sketch the trajectory of the particle. (This is the trajectory of a point on the rim of a wheel that is rolling at a constant speed on a horizontal surface. The curve traced out by such a point as it moves through space is called a cycloid.) (b) Determine the velocity components and the acceleration components of the particle at any time \(t\) . (c) At which times is the particle momentarily at rest? What are the coordinates of the particle at these times? What are the magnitude and direction of the acceleration at these times? (d) Does the magnitude of the acceleration depend on time? Compare to uniform circular motion.
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Circular Motion and Gravitation

Read Explanation

Work Energy and Power

Read Explanation

Particle Model of Matter

Read Explanation

Fundamentals of Physics

Read Explanation

Nuclear 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.