/*! 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 40 BIO Suppose a chinook salmon nee... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 40

BIO Suppose a chinook salmon needs to jump a waterfall that is \(1.50 \mathrm{~m}\) high. If the fish starts from a distance \(1.00 \mathrm{~m}\) from the base of the ledge over which the waterfall flows, (a) find the \(x\)-and \(y\)-components of the initial velocity the salmon would need to just reach the ledge at the top of its trajectory. (b) Can the fish make this jump? (Note that a chinook salmon can jump out of the water with an initial speed of \(6.26 \mathrm{~m} / \mathrm{s}\).)

Short Answer

Expert verified
To answer part (a), use the given exercise parameters and the trigonometric relationships to find the x and y-components of the velocity. For part (b), compare the found values with the maximum initial speed to determine if the jump is possible or not. The specific numerical answer depends on the parameters given.

Step by step solution

01

Determine the properties of the trajectory

First, plot the trajectory and trajectory path of the salmon. The trajectory's starting point will be the base of waterfall where the salmon begins its jump (the origin), and the ending point is the top of the waterfall. One can assume that trajectory is parabolic (like all ballistic motions). The height of the trajectory \(h\) is given to be 1.50 m (y-direction), and the horizontal distance \(d\) the salmon needs to jump is 1.00 m (x-direction).
02

Decompose the initial velocity

The goal of the initial step is to decompose the initial velocity into its x and y components. The salmon's total initial velocity (\(v_{0}\)) is said to be 6.26 m/s. To decompose this, one needs to determine the launch angle (\(胃\)), which is given by the inverse tangent of the y and x components, whom we denote as \(v_{0y}\) for y-component and \(v_{0x}\) for x-component. Thus we have: \(胃 = atan(h/d)\).
03

Calculate the initial velocities

Using the launch angle, one can now find the x and y-components of the initial velocity. The y-component is determined by: \(v_{0y} = v_0 * sin(胃)\). Similarly, the x-component is defined as: \(v_{0x} = v_0 * cos(胃)\). These equations are derived from basic trigonometric functions in consideration to vector components, where \(胃\) is the angel between the x-axis and the velocity vector.
04

Check if the salmon can make the jump

Now that the initial velocity components are calculated, we compare the values of \(v_{0y}\) and \(v_{0x}\) to the salmon's maximum allowed initial speed of 6.26 m/s to check if the jump is feasible or not. If both \(v_{0y}\) and \(v_{0x}\) are smaller or equal to 6.26 m/s, then yes it will be possible for the salmon to make the jump. Otherwise, it won'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.

Initial Velocity Components
In projectile motion, one of the foundational aspects is breaking down the initial velocity into its horizontal and vertical components. This step simplifies the analysis as it allows us to independently examine motion in the x and y directions. For this exercise involving the salmon, the total initial speed is provided as 6.26 m/s. We need to decompose this velocity to understand how much of it contributes to horizontal (x-direction) and vertical (y-direction) movements.
To achieve this, we utilize trigonometry:
  • The angle of launch, \( \theta \), is found using the relation: \( \theta = \tan^{-1}(\frac{y}{x}) \). Here, \( y \) is the vertical height (1.50 m) the salmon must jump, and \( x \) is the horizontal distance (1.00 m).
  • With \( \theta \) known, we use sine and cosine functions to find the components:
    • \( v_{0y} = v_0 \cdot \sin(\theta) \) for the vertical component.
    • \( v_{0x} = v_0 \cdot \cos(\theta) \) for the horizontal component.

Utilizing these components will help us in the subsequent analysis of the salmon's jump.
Trajectory Analysis
Next, let's analyze the trajectory of the salmon, which is idealized as a parabolic path, typical of projectile motion. Certain assumptions are made, such as neglecting air resistance, to simplify our calculations.
Here's a brief breakdown:
  • The motion in the x-direction is uniform. This means that once the salmon is airborne, its velocity in the horizontal direction remains constant as no forces act horizontally (assuming no air resistance).
  • Conversely, in the y-direction, gravity acts as a constant force, causing a change in the vertical speed over time. Its influence results in a vertical acceleration of \(-9.81\, \text{m/s}^2\).
By separating the motions in each direction, we can calculate how long the salmon remains airborne and the path it follows:
  • The time of flight can be determined by considering the vertical motion and is dependent on the initial vertical speed and the height of the waterfall.
  • With the flight time, we can also calculate how far horizontally the salmon travels, ensuring it covers the necessary 1.00 m to reach the waterfall's ledge.

This detailed exploration of the trajectory helps confirm if our initial velocity components are sufficient for the salmon to clear its target.
Maximum Jump Speed
Finally, let's discuss if the salmon can achieve its jump with its known capabilities. With a maximum initial speed of 6.26 m/s, we already decomposed this speed into the necessary components for trajectory analysis.
To check the feasibility:
  • We compare the vector sum (magnitude) of our calculated initial velocity components \(v_{0x}\) and \(v_{0y}\) with the given maximum jump speed.
  • Using the Pythagorean theorem, we find the resultant speed: \( v_0 = \sqrt{v_{0x}^2 + v_{0y}^2} \).
If this resultant speed is less than or equal to 6.26 m/s, the jump is feasible:
  • This condition checks if the salmon can use its full potential to achieve the necessary height and distance combined.
  • If the computed speed exceeds 6.26 m/s, the jump cannot be completed as planned without exceeding the salmon's jumping capability.
With these calculations, we can confidently answer whether the salmon can successfully leap over the waterfall.

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 novice golfer on the green takes three strokes to sink the ball. The succes- sive displacements of the ball are \(4.00 \mathrm{~m}\) to the north, \(2.00 \mathrm{~m}\) northeast, and \(1.00 \mathrm{~m}\) at \(30.0^{\circ}\) west of south (Fig. P3.21). Starting at the same initial point, an expert golfer could make the hole in what single displacement?

A rocket is launched at an angle of \(53.0^{\circ}\) above the horizontal with an initial speed of \(100 \mathrm{~m} / \mathrm{s}\). The rocket moves for \(3.00 \mathrm{~s}\) along its initial line of motion with an acceleration of \(30.0 \mathrm{~m} / \mathrm{s}^{2}\). At this time, its engines fail and the rocket proceeds to move as a projectile. Find (a) the maximum altitude reached by the rocket, (b) its total time of flight, and (c) its horizontal range.

A roller coaster moves \(200 \mathrm{ft}\) horizontally and then rises \(135 \mathrm{ft}\) at an angle of \(30.0^{\circ}\) above the horizontal. Next, it travels \(135 \mathrm{ft}\) at an angle of \(40.0^{\circ}\) below the horizontal. Use graphical techniques to find the roller coaster's displacement from its starting point to the end of this movement.

A student decides to measure the muzzle velocity of a pellet shot from his gun. He points the gun horizontally. He places a target on a vertical wall a distance \(x\) away from the gun. The pellet hits the target a vertical distance \(y\) below the gun. (a) Show that the position of the pellet when traveling through the air is given by \(y=A x^{2}\), where \(A\) is a constant. (b) Express the constant \(A\) in terms of the initial (muzzle) velocity and the freefall acceleration. (c) If \(x=3.00 \mathrm{~m}\) and \(y=0.210 \mathrm{~m}\), what is the initial speed of the pellet?

A projectile is launched with an initial speed of \(60.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) above the horizontal. The projectile lands on a hillside \(4.00\) s later. Neglect air friction. (a) What is the projectile's velocity at the highest point of its trajectory? (b) What is the straight-line distance from where the projectile was launched to where it hits its target?
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Physics of Motion

Read Explanation

Modern Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Work Energy and Power

Read Explanation

Scientific Method 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.