/*! 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 71 Multiple-Concept Example 4 provi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 71

Multiple-Concept Example 4 provides useful background for this problem. A diver runs horizontally with a speed of \(1.20 \mathrm{m} / \mathrm{s}\) off a platform that is \(10.0 \mathrm{m}\) above the water. What is his speed just before striking the water?

Short Answer

Expert verified
The diver's speed before impact is approximately 14.05 m/s.

Step by step solution

01

Understand the Problem

A diver runs horizontally off a platform 10 meters above water, with an initial horizontal speed of 1.20 m/s. We need to find the diver's speed just before hitting the water. This involves both horizontal and vertical motion components.
02

Analyze Horizontal Motion

In horizontal motion, the speed remains constant as there is no acceleration. Thus, the horizontal speed of the diver remains 1.20 m/s.
03

Analyze Vertical Motion

In vertical motion, the diver is under constant acceleration due to gravity (9.8 m/s虏). Using the kinematic equation for final velocity under constant acceleration, \( v^2 = u^2 + 2as \), where \( u = 0 \), \( a = 9.8 \ m/s^2 \), and \( s = 10 \ m \), we calculate the final vertical velocity just before impact.
04

Calculate Vertical Velocity

We apply \( v = \sqrt{2 \times 9.8 \times 10} \). Thus, the vertical velocity \( v \approx \sqrt{196} = 14 \ m/s \).
05

Find the Total Speed Before Impact

The total speed of the diver just before striking the water is the vector sum of his horizontal and vertical velocities. We use the Pythagorean theorem: \( v_{total} = \sqrt{(1.20)^2 + (14)^2} \).
06

Calculate Total Speed

\( v_{total} = \sqrt{1.44 + 196} = \sqrt{197.44} \approx 14.05 \ m/s \). Thus, the diver's speed just before hitting the water is approximately 14.05 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 Motion
In physics problems, when an object moves horizontally, its motion is independent of any vertical influences. This is the case when the diver runs off the platform. Here, we talk about motion parallel to the Earth's surface. The key aspect is that, without external horizontal forces acting (such as friction or air resistance for simplicity), the horizontal speed remains constant throughout the motion.

In the given exercise, the diver starts with a horizontal speed of 1.20 m/s. Since horizontal motion is constant, this speed does not change until the diver hits the water.

To summarize:
  • Horizontal speed remains constant: 1.20 m/s
  • No horizontal acceleration
  • All horizontal motion depends on initial speed
Vertical Motion
Vertical motion of an object is influenced by gravity, which causes acceleration towards the Earth's surface. This vertical component starts from rest and gains speed continuously as long as the fall continues. Gravity on Earth is approximately 9.8 m/s虏.

When the diver jumps off the platform, he begins to fall under the influence of gravity. Initially, his vertical speed is zero. We use kinematic equations to calculate how quickly he gains speed before reaching the water. Specifically, we use:
  • Initial vertical speed, \(u = 0\)
  • Acceleration due to gravity, \(a = 9.8 \, ext{m/s}^2\)
  • Displacement, \(s = 10 \, ext{m}\)
  • Final vertical speed, \(v = \sqrt{2as} = 14 \, ext{m/s}\)
Kinematic Equations
Kinematic equations are vital tools for solving motion problems without considering forces. They relate different motion parameters like velocity, acceleration, displacement, and time. We use these equations to analyze the movement of objects under constant acceleration.

In this situation, the diver's vertical motion applies the kinematic equation \( v^2 = u^2 + 2as \), useful in finding the final velocity when starting from rest:
  • \( v \) is the final velocity
  • \( u \) is the initial velocity
  • \( a \) is acceleration
  • \( s \) is the displacement
In our exercise, the calculations showed the vertical speed just before water contact as 14 m/s.
Vector Sum
The concept of vector sum is crucial when combining different directions of motion, like horizontal and vertical. This involves considering both effects simultaneously to get the total resultant velocity or speed as a vector.

Using the Pythagorean theorem, this can be calculated as:
  • Horizontal velocity, \(v_x = 1.20 \, ext{m/s}\)
  • Vertical velocity, \(v_y = 14 \, ext{m/s}\)
  • Total speed before impact, \(v_{total} = \sqrt{v_x^2 + v_y^2}\)
In our diver's case:
  • \( v_{total} = \sqrt{(1.20)^2 + (14)^2} \approx 14.05 \, ext{m/s}\)
This vector sum helps us see how different motion components interact leading up to the moment of impact.

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 spacecraft is traveling with a velocity of \(v_{0 x}=5480 \mathrm{m} / \mathrm{s}\) along the \(+x\) direction. Two engines are turned on for a time of 842 s. One engine gives the spacecraft an acceleration in the \(+x\) direction of \(a_{x}=1.20 \mathrm{m} / \mathrm{s}^{2},\) while the other gives it an acceleration in the \(+y\) direction of \(a_{y}=8.40 \mathrm{m} / \mathrm{s}^{2} .\) At the end of the firing, find (a) \(v_{x}\) and (b) \(v_{y}\)

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?

Two trees have perfectly straight trunks and are both growing perpendicular to the flat horizontal ground beneath them. The sides of the trunks that face each other are separated by \(1.3 \mathrm{m} .\) A frisky squirrel makes three jumps in rapid succession. First, he leaps from the foot of one tree to a spot that is \(1.0 \mathrm{m}\) above the ground on the other tree. Then, he jumps back to the first tree, landing on it at a spot that is \(1.7 \mathrm{m}\) above the ground. Finally, he leaps back to the other tree, now landing at a spot that is \(2.5 \mathrm{m}\) above the ground. What is the magnitude of the squirrel's displacement?

A golfer, standing on a fairway, hits a shot to a green that is elevated \(5.50 \mathrm{m}\) above the point where she is standing. If the ball leaves her club with a velocity of \(46.0 \mathrm{m} / \mathrm{s}\) at an angle of \(35.0^{\circ}\) above the ground, find the time that the ball is in the air before it hits the green.
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Classical Mechanics

Read Explanation

Physical Quantities And Units

Read Explanation

Nuclear Physics

Read Explanation

Physics of Motion

Read Explanation

Fundamentals of 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.