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A baseball is launched from the bat at an angle \(\theta_{0}=30^{\circ}\) with respect to the positive \(x\) -axis and with an initial speed of \(40 \mathrm{~m} / \mathrm{s}\), and it is caught at the same height from which it was hit. Assuming ideal projectile motion (positive \(y\) -axis upward), the velocity of the ball when it is caught is a) \((20.00 \hat{x}+34.64 \hat{y}) \mathrm{m} / \mathrm{s}\). b) \((-20.00 \hat{x}+34.64 \hat{y}) \mathrm{m} / \mathrm{s}\) c) \((34.64 \hat{x}-20.00 \hat{y}) \mathrm{m} / \mathrm{s}\) d) \((34.64 \hat{x}+20.00 \hat{y}) \mathrm{m} / \mathrm{s}\).

Short Answer

Expert verified
Short Answer: To solve this problem, we can follow these steps: 1. Calculate the initial velocity components in the x and y directions using trigonometry. 2. Calculate the time of flight when the ball reaches the same height as it was launched. 3. Calculate the final velocity components in x and y directions using equations of motion. 4. Compare the calculated components with the given options to find the correct answer. The final velocity of the baseball will be given by the components that match one of the given options.

Step by step solution

01

Calculate the Initial Velocity Components

We need to calculate the initial velocity components along the x and y direction. Since the initial speed is 40 m/s and the launch angle is \(30^{\circ}\), we can find the components using trigonometry: \(V_{0x} = V_0 \cos\theta_{0} = 40 \cos30^{\circ}\) \(V_{0y} = V_0 \sin\theta_{0} = 40 \sin30^{\circ}\)
02

Calculate the Time of Flight

To find the time of flight, we need to find out when the baseball reaches the same height it was launched from. Since there is zero initial displacement along the y-axis, we can use the following equation to find the time of flight: \(y = V_{0y}t - \frac{1}{2}gt^2\) Since the baseball is caught at the same height, \(y = 0\). Plugging the values into the above equation, we get: \(0 = V_{0y}t - \frac{1}{2}(9.8)t^2\) Solving for t, we get two possible values for the time of flight. One value will be for the time it takes to reach its peak, and the other for when it lands back to its initial height. We are interested in the landing time.
03

Calculate Final Velocity Components

We can calculate the final x and y components of the velocity by using the equations of motion: In the x-direction, the velocity remains constant: \(V_{fx} = V_{0x}\) In the y-direction, we need to account for the acceleration due to gravity: \(V_{fy} = V_{0y} - gt\) Now, we can plug in the values we have calculated earlier and find the final velocity components.
04

Compare Results with the Given Options

Now that we have calculated the final velocity components along x and y directions, we can compare them with the given options. We will find that the components match one of the given options, which will be our answer. In this way, by calculating initial velocity components, time of flight, and final velocity components, we can find the correct option for the velocity of the ball when it is caught at the same height it was hit.

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Key Concepts

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

Initial Velocity Components
Understanding the initial velocity components is fundamental in projectile motion. These components determine how a projectile moves in both the horizontal and vertical directions. To find these components, we decompose the initial speed into two parts: horizontal (\( V_{0x} \)) and vertical (\( V_{0y} \)). Given an initial speed of 40 m/s at a launch angle of \( 30^{\circ} \), you can use trigonometry:
  • For the horizontal component,\( V_{0x} = V_0 \cos \theta_{0} = 40 \cos 30^{\circ} \). This evaluates to about\( 34.64 \text{ m/s} \).
  • For the vertical component,\( V_{0y} = V_0 \sin \theta_{0} = 40 \sin 30^{\circ} \). This gives\( 20.00 \text{ m/s} \).
These calculations show how we separate the initial velocity into directional components, which are used in later calculations to predict the projectile's path and behavior.
Time of Flight
The time of flight is a crucial aspect in analyzing projectile motion. It tells us the total time that a projectile is in the air. For a baseball launched and caught at the same height, the calculation considers the entire arc's time, from launch to landing. To determine time of flight, we use the formula for vertical motion:\[ y = V_{0y}t - \frac{1}{2}gt^2 \]Since the baseball begins and ends at the same vertical position, \( y = 0 \). Substituting this in, we have:\[ 0 = V_{0y}t - \frac{1}{2}(9.8)t^2 \]When simplified, this equation presents two solutions:
  • The time ascending to its peak
  • The total flight time (when the baseball returns to starting height)
The second value is of interest, indicating the time from being hit to being caught again. This is also the time used to compute the final vertical velocity component.
Final Velocity Calculation
Calculating final velocity involves understanding how velocity changes over time due to gravity. Both horizontal and vertical components need to be considered.In projectile motion, the horizontal velocity component remains constant throughout the flight, meaning:\[ V_{fx} = V_{0x} = 34.64 \text{ m/s} \]However, gravity affects the vertical component. The formula to find the vertical velocity at any time \( t \) is:\[ V_{fy} = V_{0y} - gt \]For a baseball caught at the same height it was launched from, the vertical component of velocity will have identical magnitude but opposite direction upon landing. Thus, the final vertical velocity component is:\[ V_{fy} = - V_{0y} = -20.00 \text{ m/s} \]Combining these components provides the final velocity vector for the baseball as it returns to its original height. This is crucial for analyzing the overall motion and matches with option b) in the original problem statement: \((-20.00 \hat{x} + 34.64 \hat{y}) \text{ m/s}\).

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Most popular questions from this chapter

At the end of the spring term, a high school physics class celebrates by shooting a bundle of exam papers into the town landfill with a homemade catapult. They aim for a point that is \(30.0 \mathrm{~m}\) away and at the same height from which the catapult releases the bundle. The initial horizontal velocity component is \(3.90 \mathrm{~m} / \mathrm{s}\). What is the initial velocity component in the vertical direction? What is the launch angle?

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