/*! 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 48 Martian Athletics. In the long j... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Martian Athletics. In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the greatest horizontal distance. Suppose that on earth she is in the air for time \(T\) , reaches a maximum height \(h,\) and achieves a horizontal distance \(D .\) If she jumped in exactly the same way during a competition on Mars, where \(g_{\text {Mars}}\) is 0.379 of its earth value, find her time in the air, maximum height, and horizontal distance. Express each of these three quantities in terms of its earth value. Air resistance can be neglected on both planets.

Short Answer

Expert verified
On Mars: \(T_{\text{Mars}} = \frac{T}{0.379}\), \(h_{\text{Mars}} = \frac{h}{0.379}\), \(D_{\text{Mars}} = \frac{D}{0.379}\).

Step by step solution

01

Understanding the Problem

The athlete's jump on Earth has time in the air \(T\), a maximum height \(h\), and a horizontal distance \(D\). We need to determine how these values change on Mars, where gravity \(g_{\text{Mars}}\) is 0.379 times the Earth's gravity.
02

Analyzing Time in the Air

The time the athlete is in the air, \(T\), is determined by the vertical motion. On Earth, the time of flight \(T\) is related to gravity and initial vertical velocity. Since \(T = \frac{2v_{0y}}{g_{\text{Earth}}}\), where \(v_{0y}\) is the initial vertical velocity, the time on Mars \(T_{\text{Mars}} = \frac{2v_{0y}}{g_{\text{Mars}}}\). Considering \(g_{\text{Mars}} = 0.379g_{\text{Earth}}\), \(T_{\text{Mars}} = \frac{T}{0.379}\).
03

Calculating Maximum Height

The maximum height \(h\) is achieved when the vertical velocity is zero. Using \(h = \frac{v_{0y}^2}{2g_{\text{Earth}}}\), for Mars, the height \(h_{\text{Mars}} = \frac{v_{0y}^2}{2g_{\text{Mars}}}\). Since \(g_{\text{Mars}} = 0.379g_{\text{Earth}}\), we find \(h_{\text{Mars}} = \frac{h}{0.379}\).
04

Determining Horizontal Distance

The horizontal distance \(D\) depends on time in the air and initial horizontal speed. Since horizontal velocity \(v_{0x}\) remains unchanged between the two planets, \(D_{\text{Mars}} = v_{0x} T_{\text{Mars}}\). Therefore, \(D_{\text{Mars}} = D \times \frac{1}{0.379}\).
05

Conclusion

On Mars, the time in the air \(T_{\text{Mars}} = \frac{T}{0.379}\), the maximum height \(h_{\text{Mars}} = \frac{h}{0.379}\), and the horizontal distance \(D_{\text{Mars}} = \frac{D}{0.379}\). Thus, all these values are greater by the factor of \(\frac{1}{0.379}\) compared to Earth.

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 91Ó°ÊÓ!

Key Concepts

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

Kinematics
Kinematics is a branch of physics that describes the motion of objects without considering the causes of motion. It focuses on parameters like displacement, velocity, acceleration, and time to describe motion. In our Martian athletics problem, kinematics plays a crucial role as we analyze the trajectory of an athlete's long jump.
Understanding the underlying kinematic equations helps in predicting how motion changes in different environments. On Earth, for instance, the time of flight, maximum height, and horizontal distance of an athlete's jump can be explained using kinematics principles. These components depend on initial velocity, the angle of launch, and gravitational force. The equations tell us how each of these factors influences the overall motion.
In the context of our problem, kinematics allows us to explore how these factors change when the environment shifts from Earth to Mars, where gravity differs significantly. Understanding these changes through kinematic equations helps us adapt our problem-solving strategy efficiently.
Gravity on Mars
Gravity on Mars is a fraction of Earth's gravity, specifically around 0.379 times weaker. This change in gravitational force significantly affects an athlete's jumping performance. On Earth, gravity (approximately 9.81 m/s extsuperscript{2}) pulls objects down more strongly than on Mars.
When the athlete jumps in a Martian environment, the reduced gravitational pull leads to longer time in the air and greater maximum height achieved by the athlete. The reason for this is that less force means the object takes longer to return to the ground. This slower descent also provides more time for horizontal travel, influencing the distance covered during the jump.
Understanding this concept is vital in physics problem-solving as it illustrates how environmental factors can change the dynamics of motion. In our scenario, computing the jump's time in the air, height, and distance requires accounting for Mars's gravity influence and adjusting formulas accordingly.
Vertical and Horizontal Components
In physics, breaking down motion into vertical and horizontal components simplifies the analysis of projectile motion. The vertical component involves calculating factors influenced by gravity, while the horizontal component remains unaffected by gravity in ideal conditions.
For our problem, on Earth, the vertical component helps determine the jump's time in the air and maximum height using formulas like \[ T = \frac{2v_{0y}}{g} \] and \[ h = \frac{v_{0y}^2}{2g} \] where \(v_{0y}\) is the initial vertical velocity and \(g\) represents gravity. Meanwhile, the horizontal component determines the distance traveled and remains constant between planets since no horizontal force acts on the athlete once airborne, expressed as \[ D = v_{0x} \times T \].
On Mars, although vertical motion changes due to lower gravity, the horizontal velocity component remains the same. Hence, adapting these components to Mars's environment means applying the reduced gravitational acceleration to obtain the new values for time, height, and distance.
Physics Problem Solving
Physics problem-solving involves a step-by-step approach to dissect and solve challenges involving physical concepts. The key is to clearly understand the problem, identify given information, and apply the appropriate physics principles.
In our Martian athletics problem, the first step is understanding how the change in environmental conditions, such as differing gravity, affects motion. From there, translating this understanding into equations allows us to predict the athlete's performance.
These solutions involve two main adaptations: adjusting the time of flight calculation, \(T_{\text{Mars}} = \frac{T}{0.379}\), recalibrating maximum height, \(h_{\text{Mars}} = \frac{h}{0.379}\), and recalculating the horizontal distance, \(D_{\text{Mars}} = \frac{D}{0.379}\) from their Earth values. This problem-solving process not only provides precise answers but also deepens our understanding of how physics principles apply in real-world scenarios across different environments.
  • Identify known and unknown variables.
  • Calculate each component step-by-step.
  • Adapt equations for new environments.
By practicing such methods, students build a habit of logical thinking and analytical skills necessary for tackling various physics problems.

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 jet plane is flying at a constant altitude. At time \(t_{1}=0\) it has components of velocity \(v_{x}=90 \mathrm{m} / \mathrm{s}, v_{y}=110 \mathrm{m} / \mathrm{s}\) . At time \(t_{2}=30.0 \mathrm{s}\) the components are \(v_{x}=-170 \mathrm{m} / \mathrm{s}, v_{y}=40 \mathrm{m} / \mathrm{s}\) (a) Sketch the velocity vectors at \(t_{1}\) and \(t_{2} .\) How do these two vectors differ? For this time interval calculate (b) the components of the average acceleration, and (c) the magnitude and direction of the average acceleration.

Firemen are shooting a stream of water at a burning building using a high- pressure hose that shoots out the water with a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) as it leaves the end of the hose. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation \(\alpha\) of the hose until the water takes 3.00 s to reach a building 45.0 m away. You can ignore air resistance; assume that the end of the hose is at ground level. (a) Find the angle of elevation \(\alpha\) . (b) Find the speed and acceleration of the water at the highest point in its trajectory. (c) How high above the ground does the water strike the building, and how fast is it moving just before it hits the building?

The position of a squirrel running in a park is given by \(\vec{r}=\left[(0.280 \mathrm{m} / \mathrm{s}) t+\left(0.0360 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}\right] \hat{\imath}+\left(0.0190 \mathrm{m} / \mathrm{s}^{3}\right) t^{3} \hat{\jmath}\) . (a) What are \(v_{x}(t)\) and \(v_{y}(t),\) the \(x\) - and \(y\) -components of the velocity of the squirrel, as functions of time? (b) At \(t=5.00 \mathrm{s}\) how far is the squirrel from its initial position? ( c) At \(t=5.00 \mathrm{s}\) what are the magnitude and direction of the squirrel's velocity?

A 124 -kg balloon carrying a 22 -kg basket is descending with a constant downward velocity of 20.0 \(\mathrm{m} / \mathrm{s} . \mathrm{A} 1.0\) -kg stone is thrown from the basket with an initial velocity of 15.0 \(\mathrm{m} / \mathrm{s}\) perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. The person in the basket sees the stone hit the ground 6.00 s after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 \(\mathrm{m} / \mathrm{s} .\) (a) How high was the balloon when the rock was thrown out? (b) How high is the balloon when the rock hits the ground? (c) At the instant the rock hits the ground, how far is it from the basket? (d) Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer (i) at rest in the basket and (ii) at rest on the ground.

The earth has a radius of 6380 \(\mathrm{km}\) and turns around once on its axis in 24 \(\mathrm{h}\) . (a) What is the radial acceleration of an object at the earth's equator? Give your answer in \(\mathrm{m} / \mathrm{s}^{2}\) and as a fraction of \(g .\) (b) If \(a_{\text { rad }}\) at the equator is greater than \(g,\) objects will fly off the earth's surface and into space. (We will see the reason for this in Chapter 5.) What would the period of the earth's rotation have to be for this to occur?

See all solutions

Recommended explanations on Physics Textbooks

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.