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When chasing a hare along a flat stretch of ground, a greyhound leaps into the air at a speed of \(10.0 \mathrm{m} / \mathrm{s},\) at an angle of \(31.0^{\circ}\) 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 8.77 meters and the time in the air is about 1.06 seconds.

Step by step solution

01

Break Down Velocity Components

We start by finding the horizontal and vertical components of the initial velocity. The horizontal component \(v_{x}\) is given by \(v_x = v \cdot \cos(\theta)\), and the vertical component \(v_{y}\) is given by \(v_y = v \cdot \sin(\theta)\), where \(v = 10.0 \, \text{m/s}\) and \(\theta = 31.0^{\circ}\). Calculate these as follows: \(v_x = 10.0 \, \text{m/s} \cdot \cos(31.0^{\circ})\) and \(v_y = 10.0 \, \text{m/s} \cdot \sin(31.0^{\circ})\).
02

Calculate Time of Flight

The time the greyhound is in the air can be determined using the vertical motion. Since the vertical velocity will be zero at the peak, and symmetry suggests the upward and downward times are equal, the total time \(T\) in the air can be calculated from the formula: \( T = \frac{2 \cdot v_y}{g} \) where \( g = 9.81 \, \text{m/s}^2\) is the acceleration due to gravity.
03

Calculate Horizontal Range

The range \(R\) of the leap is determined by the horizontal velocity component and the total time in the air. Use the formula: \( R = v_x \cdot T \), substituting the horizontal velocity from Step 1 and the time of flight from Step 2.

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Key Concepts

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

Velocity Components
In projectile motion, understanding the velocity components is foundational. When an object is projected at an angle, the initial velocity can be split into two parts: the horizontal component and the vertical component. These components help us analyze and predict the object's path.

- **Horizontal Component**: The horizontal component of velocity, denoted as \(v_x\), affects how far the object travels horizontally. It is calculated using the formula \(v_x = v \cdot \cos(\theta)\), where \(v\) is the launch speed and \(\theta\) is the launch angle. It remains constant throughout the projectile motion because no external forces act horizontally.- **Vertical Component**: The vertical component, \(v_y\), influences the height and time the object stays in the air. It's determined by \(v_y = v \cdot \sin(\theta)\). Unlike \(v_x\), this component changes due to the force of gravity, which pulls the object downward. By understanding these components, you can deconstruct any projectile's behavior in both the horizontal and vertical planes.
Time of Flight
The time of flight refers to the total duration a projectile remains in the air from launch until it returns to the ground. Understanding how to calculate this time is crucial for solving many projectile motion problems.

The time of flight is primarily dictated by the vertical component of the initial velocity and the influence of gravity. Since the projectile takes equal time to ascend to its peak and descend back to the ground due to symmetry of motion, we can use the time it takes to reach the highest point to find the total time.

The formula for the time of flight, \( T = \frac{2 \cdot v_y}{g} \), accounts for both ascent and descent times. Here, \(g\) stands for acceleration due to gravity, approximately \(9.81 \, \text{m/s}^2\). This formula results from the fact that it takes twice the time to ascend to the peak to complete the entire journey. Mastering this will allow you to solve for the time duration in any projectile motion scenario.
Horizontal Range
The horizontal range of a projectile motion is the total horizontal distance traveled during the flight. It's a key measure of how far the object can go and depends on both the horizontal velocity component and the time of flight.

To determine the range, use the formula \( R = v_x \cdot T \). This calculation involves multiplying the constant horizontal velocity \(v_x\) by the total time in the air \(T\), derived from the earlier time of flight concept.

Important aspects influencing the range include:
  • **Initial Speed**: Higher speeds lead to greater ranges.
  • **Launch Angle**: The angle impacts both velocity components, and optimally, \(45^\circ\) provides the maximum range in the absence of air resistance.
  • **Gravity**: Influences the vertical motion, indirectly affecting the range.
By comprehensively understanding these factors, predicting the distance of a projectile can be intuitive and accurate.

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Most popular questions from this chapter

In a mall, a shopper rides up an escalator between floors. At the top of the escalator, the shopper turns right and walks \(9.00 \mathrm{m}\) to a store. The magnitude of the shopper's displacement from the bottom of the escalator to the store is \(16.0 \mathrm{m}\). The vertical distance between the floors is \(6.00 \mathrm{m}\). At what angle is the escalator inclined above the horizontal?

A golfer imparts a speed of \(30.3 \mathrm{m} / \mathrm{s}\) to a ball, and it travels the maximum possible distance before landing on the green. The tee and the green are at the same elevation. (a) How much time does the ball spend in the air? (b) What is the longest hole in one that the golfer can make, if the ball does not roll when it hits the green?

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