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Chapter 3: Problem 70

When chasing a hare along a flat stretch of ground, a greyhound leaps into the air at a speed of 10.0 m/s, at an angle of 31.0 above the horizontal. (a) What is the range of his leap and (b) for how much time is he in the air?

Short Answer

Expert verified
The range of the leap is approximately 8.98 meters, and the time in the air is about 1.05 seconds.

Step by step solution

01

Break Down Components of Velocity

The greyhound's initial speed is 10.0 m/s at an angle of 31.0 degrees. We need to find the horizontal (x) and vertical (y) components of this velocity. Use the equations:\[ v_x = v \cos(\theta) \] and \[ v_y = v \sin(\theta) \]where \( v = 10.0 \text{ m/s} \) and \( \theta = 31.0 \text{ degrees} \).
02

Calculate Horizontal and Vertical Components

Substitute the values into the equations:\[ v_x = 10.0 \cos(31.0) \approx 8.55 \text{ m/s} \]\[ v_y = 10.0 \sin(31.0) \approx 5.15 \text{ m/s} \]
03

Determine Time in Air

To find the time the greyhound is in the air, use the vertical motion equation. Since the greyhound lands at the same vertical level from which it started, we have:\[ t = \frac{2v_y}{g} \]where \( g = 9.81 \text{ m/s}^2 \) is the acceleration due to gravity. Substitute the value of \( v_y \):\[ t = \frac{2 \times 5.15}{9.81} \approx 1.05 \text{ s} \]
04

Calculate the Range of the Leap

The range can be found using the horizontal motion. The formula is:\[ \text{Range} = v_x \times t \]Substituting the known values:\[ \text{Range} = 8.55 \times 1.05 \approx 8.98 \text{ m} \]
05

Conclusion and Summary

The greyhound's leap covers a horizontal distance of approximately 8.98 meters and the time of flight is approximately 1.05 seconds. These values were calculated using the component velocities and the equations of motion for projectile trajectories.

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Key Concepts

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

Kinematic Equations
Kinematic equations are foundational in understanding projectile motion. They describe the motion of objects under the influence of gravity, neglecting air resistance. In projectile problems, different equations are used to analyze horizontal and vertical motion separately.

Here are the kinematic equations you'll use:
  • Vertical motion: The time of flight can be found using the equation \[ t = \frac{2v_y}{g} \], where \( v_y \) is the initial vertical velocity component, and \( g \) is the acceleration due to gravity (9.81 m/s虏).
  • Horizontal motion: The range of the projectile is given by \[ \text{Range} = v_x \times t \], where \( v_x \) is the horizontal velocity component, and \( t \) is the total time in the air.
The synergy between these equations helps you solve for various parameters like time, range, and velocity in projectile motion. Each axis (horizontal and vertical) is treated independently to simplify calculations.
Vector Components
Understanding vector components is crucial for breaking down projectile motion. Here, the greyhound's leap can be divided into horizontal and vertical components using trigonometry.

To find these components, use:
  • Horizontal component: \[ v_x = v \cos(\theta) \]
    Here, \( v \) is the initial speed, and \( \theta \) is the angle of launch (31.0 degrees). For the greyhound, this results in \( v_x \approx 8.55 \text{ m/s} \).
  • Vertical component: \[ v_y = v \sin(\theta) \]
    This gives the upward speed at launch, calculated as \( v_y \approx 5.15 \text{ m/s} \).
These components are essential for analyzing the motion separately in horizontal and vertical directions. Knowing how to resolve velocity into components through cosine and sine functions streamlines the process of solving projectile motion problems.
Time of Flight
The time of flight refers to how long the greyhound is in the air. To determine this, consider the vertical motion. Since the greyhound leaps and lands at the same level, the formula \[ t = \frac{2v_y}{g} \] applies.

Inserting the vertical velocity \( v_y \) and gravity \( g = 9.81 \text{ m/s}^2 \), you find:
  • \[ t = \frac{2 \times 5.15}{9.81} \approx 1.05 \text{ seconds} \]
This tells you that the greyhound remains airborne for approximately 1.05 seconds. Accurately calculating time in the air is key for further computations, such as determining the range. The symmetric parabolic trajectory of projectiles means the time ascending equals the time descending.
Range Calculation
The range of a projectile is the horizontal distance it covers during its flight. For the greyhound's leap, we focus on the horizontal velocity component and time of flight.

Using the equation \[ \text{Range} = v_x \times t \], you plug in:
  • \( v_x = 8.55 \text{ m/s} \) (horizontal speed)
  • \( t = 1.05 \text{ seconds} \) (time in air)
The calculation becomes:
  • \[ \text{Range} = 8.55 \times 1.05 \approx 8.98 \text{ meters} \]
The greyhound's leap spans about 8.98 meters. Understanding range calculation requires correctly determining both the horizontal velocity and the flight duration. These components interact to produce the total distance traveled horizontally.

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