/*! 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 21 The froghopper, Philaenus spumar... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 21

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of \(58.0^{\circ}\) above the horizontal, some of the tiny critters have reached a maximum height of \(58.7 \mathrm{~cm}\) above the level ground. (See Nature, Vol. 424,31 July \(2003,\) p. \(509 .\) ) (a) What was the takeoff speed for such a leap? (b) What horizontal distance did the froghopper cover for this world-record leap?

Short Answer

Expert verified
Takeoff speed: 3.98 m/s, horizontal distance: 1.46 m.

Step by step solution

01

Understand the problem

The problem involves finding the takeoff speed and the horizontal distance of a jump made by the froghopper. Given are the angle of projection and the maximum height reached.
02

Analyze the motion components

The jump can be broken down into horizontal and vertical components. We know that the maximum height reached by the froghopper is 58.7 cm, and it was achieved at a launch angle of 58.0掳.
03

Use kinematic equations for vertical motion

Apply the kinematic equation for the vertical motion to find the initial vertical velocity component:\[ v_{y}^2 = v_{0y}^2 - 2g h \]Here, \(v_{y} = 0\) at maximum height, \(h = 0.587\, \text{m}\), and \(g \approx 9.81 \, \text{m/s}^2\). Solve for \(v_{0y}\), the initial vertical velocity:\[ 0 = v_{0y}^2 - 2(9.81)(0.587) \]
04

Solve for the initial vertical velocity component

Simplify and solve the equation for \(v_{0y}\):\[ v_{0y} = \sqrt{2 \times 9.81 \times 0.587} \approx 3.39\, \text{m/s} \]
05

Use angle to find total initial velocity

Use the angle to relate total initial velocity and its vertical component:\[ v_{0y} = v_{0} \, \sin(58.0^{\circ}) \]Thus,\[ 3.39 = v_{0} \, \sin(58.0^{\circ}) \]Solve for \(v_{0}\):\[ v_{0} = \frac{3.39}{\sin(58.0^{\circ})} \approx 3.98\, \text{m/s} \]
06

Determine horizontal distance using projectile motion

Calculate the horizontal component of the initial velocity using:\[ v_{0x} = v_{0} \, \cos(58.0^{\circ}) \]\[ v_{0x} = 3.98 \, \cos(58.0^{\circ}) \approx 2.11\, \text{m/s} \]Use the time of flight and horizontal velocity to find the distance traveled. Time of flight, when starting and ending at same vertical level:\[ t = \frac{2v_{0y}}{g} = \frac{2 \times 3.39}{9.81} \approx 0.69\, \text{s} \]Finally, calculate the range (horizontal distance):\[ d = v_{0x} \times t \approx 2.11 \times 0.69 \approx 1.46\, \text{m} \]
07

Conclusion

The takeoff speed for the froghopper's leap was approximately 3.98 m/s, and the horizontal distance covered was about 1.46 meters.

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.

Kinematic Equations
Kinematic equations are essential tools used in physics to describe motion. These equations relate several key variables: displacement, initial and final velocities, acceleration, and time.
For projectile motion, kinematic equations can help determine how objects move through space. Particularly, these equations decompose complex motions into simpler horizontal and vertical components.
  • In vertical motion, gravity plays a crucial role as the primary force acting on the object.
  • For horizontal motion, assuming no air resistance, the velocity remains constant.
Understanding these equations will help in calculating factors like takeoff speed, time of flight, and distance covered.
Vertical Motion
Vertical motion in projectile motion involves analyzing how an object moves up and down under the influence of gravity. For the froghopper's jump:
  • The motion stops (velocity = 0) when reaching maximum height.
  • We use the kinematic equation: \[ v_{y}^2 = v_{0y}^2 - 2g h \] to calculate the initial vertical velocity.

Gravity accelerates the object downward at approximately 9.81 m/s虏. At the jump's peak, the upward velocity is zero, and gravity takes over entirely. By solving the equation, one can find how the froghopper achieves its remarkable height.
Horizontal Motion
While the froghopper is airborne, it also moves horizontally. In projectile motion, horizontal and vertical motions are independent of each other.
  • Since horizontal motion usually assumes no air resistance, velocity across stays constant.
  • To calculate the horizontal distance traveled, you multiply the horizontal velocity by the time of flight.
For instance, in this problem, the horizontal component is computed through trigonometric functions using the angle of launch. The range is then found using these values and the duration of the entire motion.
Launch Angle
The launch angle is critical in determining the trajectory of any projectile's motion. In the case of the froghopper, the jump occurs at an impressive 58 degrees above horizontal.
Here's why the angle matters:
  • It affects both the height and the distance covered.
  • A steeper angle (closer to 90掳) emphasizes height over range, while a shallower angle favors distance.

Trigonometry links the launch angle to initial velocity components, allowing us to calculate both vertical and horizontal speeds, which govern overall motion.
Maximum Height
Achieving maximum height in projectile motion means reaching the highest point possible. For vertical motion, it鈥檚 where the vertical velocity drops to zero, before the object begins descending.
For the froghopper, this involves intricate balance between speed and launch angle.
  • Using the kinematic equation, you can determine height by isolating the vertical component of velocity.
  • The maximum height provides insight into the energy and efficiency of the jump.
This aspect plays a crucial role in comprehending how forces like gravity compete with initial kinetic energy to set the pinnacle of projectile paths.

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 large number of seeds are observed, and their initial launch angles are recorded. The range of projection angles is found to be \(-51^{\circ}\) to \(75^{\circ},\) with a mean of \(31^{\circ} .\) Approximately \(65 \%\) of the seeds were launched between \(6^{\circ}\) and \(56^{\circ} .\) Which of the following hypotheses is best supported by these data? Seeds are preferentially launched A. at angles that maximize the height they travel above the plant. B. at angles below the horizontal in order to drive the seeds into the ground with more force. C. at angles that maximize the horizontal distance traveled from the plant. D. at angles that minimize the time the seeds spend exposed to the air.

A football is thrown with an initial upward velocity component of \(15.0 \mathrm{~m} / \mathrm{s}\) and a horizontal velocity component of \(18.0 \mathrm{~m} / \mathrm{s}\). (a) How much time is required for the football to reach the highest point in its trajectory? (b) How high does it get above its release point? (c) How much time after it is thrown does it take to return to its original height? How does this time compare with what you calculated in part (a)? Is your answer reasonable? (d) How far has the football traveled horizontally from its original position?

A player kicks a football at an angle of \(40.0^{\circ}\) from the horizontal, with an initial speed of \(12.0 \mathrm{~m} / \mathrm{s}\). A second player standing at a distance of \(30.0 \mathrm{~m}\) from the first (in the direction of the kick) starts running to meet the ball at the instant it is kicked. How fast must he run in order to catch the ball just before it hits the ground?

It is common to see birds of prey rising upward on thermals. The paths they take may be spiral-like. You can model the spiral motion as uniform circular motion combined with a constant upward velocity. Assume a bird completes a circle of radius \(8.00 \mathrm{~m}\) every \(5.00 \mathrm{~s}\) and rises vertically at a rate of \(3.00 \mathrm{~m} / \mathrm{s}\). Determine: (a) the speed of the bird relative to the ground; (b) the bird's acceleration (magnitude and direction); and (c) the angle between the bird's velocity vector and the horizontal.

A canoe has a velocity of \(0.40 \mathrm{~m} / \mathrm{s}\) southeast relative to the earth. The canoe is on a river that is flowing \(0.50 \mathrm{~m} / \mathrm{s}\) east relative to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Kinematics Physics

Read Explanation

Scientific Method Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Magnetism

Read Explanation

Engineering 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.