/*! 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 80 A baseball player hits a home ru... [FREE SOLUTION] | 91Ó°ÊÓ

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A baseball player hits a home run, and the ball lands in the left-field seats, \(7.5 \mathrm{m}\) above the point at which it was hit. It lands with a velocity of \(36 \mathrm{m} / \mathrm{s}\) at an angle of \(28^{\circ}\) below the horizontal (see the drawing). The positive directions are upward and to the right in the drawing. Ignoring air resistance, find the magnitude and direction of the initial velocity with which the ball leaves the bat.

Short Answer

Expert verified
The initial velocity is approximately 40.13 m/s at an angle of 37.2° above the horizontal.

Step by step solution

01

Analyze the Final Velocity

We know that the final velocity of the ball is \(36 \ \mathrm{m/s}\) at an angle of \(28^{\circ}\) below the horizontal. Therefore, we can resolve this velocity into horizontal \((v_{f_x})\) and vertical \((v_{f_y})\) components using the trigonometric functions cosine and sine respectively. \[ v_{f_x} = 36 \cos(28^{\circ}) \approx 31.79 \ \mathrm{m/s} \] \[ v_{f_y} = 36 \sin(28^{\circ}) \approx -16.91 \ \mathrm{m/s} \] Note the negative sign for \(v_{f_y}\) as the ball is moving downwards at that point.
02

Calculate the Time of Flight

Using the vertical motion equation that involves displacement, we can solve for the time of flight. The vertical motion equation is:\[ v_{f_y} = v_{0_y} - g t \]The displacement \(y = 7.5 \ \mathrm{m}\). We assume initial vertical position is \(0\) and upward is positive: \[ y = v_{0_y} t - \frac{1}{2} g t^2 \]Substitute \(v_{0_y} = v_{f_y} + gt\):\[ 7.5 = (-16.91) t + \frac{1}{2} \cdot 9.81 \cdot t^2 \]Solving this quadratic equation gives us: \[ t \approx 4.19 \ \mathrm{s} \]
03

Calculate the Initial Velocity Components

Now we use the time *t* to find the initial vertical velocity component \(v_{0_y}\), by rearranging the vertical motion equation:\[ v_{f_y} = v_{0_y} - g t \]Substituting the known values:\[ -16.91 = v_{0_y} - 9.81 \times 4.19 \] Solving for \(v_{0_y}\), we find:\[ v_{0_y} \approx 24.21 \ \mathrm{m/s} \]The horizontal velocity component \(v_{0_x}\) remains constant so:\[ v_{0_x} = v_{f_x} = 31.79 \ \mathrm{m/s} \]
04

Determine the Magnitude of the Initial Velocity

The magnitude of the initial velocity \(v_0\) can be found using the Pythagorean theorem:\[ v_0 = \sqrt{v_{0_x}^2 + v_{0_y}^2} \]Substitute the values:\[ v_0 = \sqrt{(31.79)^2 + (24.21)^2} \approx 40.13 \ \mathrm{m/s} \]
05

Determine the Direction of the Initial Velocity

The direction \( \theta \) of the initial velocity is given by:\[ \theta = \tan^{-1}\left(\frac{v_{0_y}}{v_{0_x}}\right) \]Substitute the known values:\[ \theta = \tan^{-1} \left(\frac{24.21}{31.79}\right) \approx 37.2^{\circ} \]This angle is measured above the horizontal.

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Key Concepts

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

Initial velocity calculation
The calculation of initial velocity is crucial in projectile motion as it determines how far and high the projectile will travel. To find the initial velocity, we must assess both its magnitude and direction. The process often begins with assessing the final velocity of a projectile. For example, in the baseball problem, the final velocity is known to be 36 m/s at an angle of 28° below the horizontal. By resolving this final velocity into horizontal and vertical components using trigonometric functions and after solving for time, we can determine the initial horizontal (\(v_{0_x}\)) and vertical velocities (\(v_{0_y}\)). Here, the horizontal component remains constant throughout the flight, meaning the initial and final horizontal velocities are equal.Once the components are known, the magnitude of the initial velocity can be calculated using the Pythagorean theorem, which combines the horizontal and vertical components into a single value, representing the speed at which the ball leaves the bat. Also, the angle of release can be found using trigonometric identities, such as the inverse tangent.
Trigonometry in physics
Trigonometry is essential in decomposing a projectile's motion into horizontal and vertical components. In physics problems like the baseball scenario, resolving the velocity at an angle often requires using sine and cosine functions. These functions help in finding the exact values of velocity in different directions, essential for accurately calculating projectile trajectories.- The cosine function is applied to find horizontal components of motion: \( v_x = v \cos(\theta) \).- The sine function helps determine the vertical components: \( v_y = v \sin(\theta) \).- Corporating angle measurements is fundamental, whether the object is moving upwards or downwards, impacting how these components are defined and calculated.Understanding these trigonometric principles allows students to solve a range of projectile motion problems by breaking down complex motion into simpler parts. This knowledge is not just pivotal in theoretical physics but is also applied widely in real-world situations.
Quadratic equations in physics
Quadratic equations often arise in physics, especially when dealing with motion problems involving time and displacement, like projectile motion. In the baseball problem, the vertical displacement (where it's hit to where it lands) can be translated into a quadratic equation to find the time of flight.Consider the basic formula for vertical motion: \( y = v_{0_y}t + \frac{1}{2} a t^2 \). Here, \( y \) is vertical displacement, \( v_{0_y} \) is initial vertical velocity, \( t \) is the time of flight, and \( a \) is acceleration due to gravity.- By substituting known values into this equation, we can find \( t \), the time variable.- Solving the resulting quadratic equation often requires rearrangement into a standard form (\( ax^2 + bx + c = 0 \)) and applying quadratic formula or factorization methods.This equation is vital in deriving other quantities related to projectile motion, enabling us to comprehensively analyze the behavior of moving projects throughout their trajectory. Understanding how to manipulate and solve these equations is crucial for students aiming to master projectile physics.

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Most popular questions from this chapter

In a marathon race Chad is out in front, running due north at a speed of \(4.00 \mathrm{m} / \mathrm{s} .\) John is \(95 \mathrm{m}\) behind him, running due north at a speed of 4.50 m/s. How long does it take for John to pass Chad?

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?

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