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Chapter 3: Problem 17

A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 m/s and at an angle of 36.9\(^\circ\) above the horizontal. Ignore air resistance. (a) At what \(two\) times is the baseball at a height of 10.0 m above the point at which it left the bat? (b) Calculate the horizontal and vertical components of the baseball's velocity at each of the two times calculated in part (a). (c) What are the magnitude and direction of the baseball's velocity when it returns to the level at which it left the bat?

Short Answer

Expert verified
(a) 0.68 s and 3.0 s. (b) At 0.68 s: \(v_x = 24.0\,\text{m/s}\), \(v_y = 11.3\,\text{m/s}\); at 3.0 s: \(v_x = 24.0\,\text{m/s}\), \(v_y = -11.4\,\text{m/s}\). (c) Velocity magnitude 30.0 m/s, direction -36.9掳.

Step by step solution

01

Decompose the Initial Velocity

The initial velocity of the baseball is given as 30.0 m/s at an angle of 36.9掳 above the horizontal. To find the horizontal and vertical components of the velocity, we use trigonometry. The horizontal component \( v_{0x} \) is given by \( v_0 \cos(36.9掳) \) and the vertical component \( v_{0y} \) is given by \( v_0 \sin(36.9掳) \).\[v_{0x} = 30.0 \cos(36.9掳) \approx 24.0\,\text{m/s}\]\[v_{0y} = 30.0 \sin(36.9掳) \approx 18.0\,\text{m/s}\]
02

Use Kinematic Equation for Vertical Motion

To find the times when the baseball is at a height of 10 meters, we use the vertical motion equation: \[h = v_{0y} t - \frac{1}{2} g t^2\]where \( h = 10.0 \) m and \( g = 9.8 \text{ m/s}^2 \). Substitute \( v_{0y} = 18.0 \text{ m/s} \) into the equation and solve for \( t \): \[10 = 18t - 4.9t^2\]
03

Solve the Quadratic Equation

Rearrange the equation from Step 2 into standard form and solve the quadratic:\[4.9t^2 - 18t + 10 = 0\]Using the quadratic formula, \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4.9 \), \( b = -18 \), and \( c = 10 \):\[t = \frac{18 \pm \sqrt{(-18)^2 - 4 \times 4.9 \times 10}}{9.8}\]\[t = \frac{18 \pm \sqrt{324 - 196}}{9.8}\]\[t = \frac{18 \pm \sqrt{128}}{9.8}\]Calculate the two time solutions:
04

Calculate Times

Solve the equation from Step 3 to find the times.\[t = \frac{18 + 11.31}{9.8} \approx 3.0 \text{ s}\]\[t = \frac{18 - 11.31}{9.8} \approx 0.68 \text{ s}\]So, the baseball is at 10.0 m at approximately 0.68 s and 3.0 s.
05

Calculate Velocity Components at Two Times

For each time, calculate the horizontal and vertical velocity components. The horizontal velocity remains constant:\[v_x = v_{0x} = 24.0\,\text{m/s}\]For the vertical velocity, use the equation \( v = v_{0y} - gt \).For \( t = 0.68 \) s:\[v_y = 18.0 - 9.8 \times 0.68 \approx 11.3\,\text{m/s}\]For \( t = 3.0 \) s:\[v_y = 18.0 - 9.8 \times 3.0 \approx -11.4\,\text{m/s}\]
06

Calculate Magnitude and Direction of Final Velocity

Use the velocities calculated at \( t = 3.0 \) seconds when the baseball is returning to the original level.For horizontal velocity:\[v_x = 24.0\,\text{m/s}\]For vertical velocity (at same level):\[v_y = - v_{0y} = -18.0\,\text{m/s}\]Calculate magnitude:\[v = \sqrt{v_x^2 + v_y^2} = \sqrt{24.0^2 + (-18.0)^2} = 30.0\,\text{m/s}\]Calculate direction \( \theta \):\[\theta = \tan^{-1} \left(\frac{-18.0}{24.0}\right) \approx -36.9掳\]This is below the horizontal.

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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 essential tools for analyzing motion, especially in physics problems like projectile motion. They provide a way to relate different kinematic variables such as displacement, velocity, acceleration, and time. In this scenario, the baseball is subject to vertical and horizontal motion that can be described using these equations.

When dealing with projectile motion, two primary kinematic equations are utilized:
  • Vertical motion: \[ h = v_{0y} t - \frac{1}{2} g t^2 \]where \( h \) is the height, \( v_{0y} \) is the initial vertical velocity, \( g \) is the acceleration due to gravity, and \( t \) is the time elapsed.
  • Vertical velocity: \[ v_y = v_{0y} - gt \]Where \( v_y \) is the vertical component of the velocity at any time \( t \).
In the context of the baseball exercise, we use these equations to find the times when it is at a certain height, solve for line-of-flight parameters, and analyze its motion.
Velocity Components
Breaking down a projectile's velocity into vertical and horizontal components is fundamental for understanding its motion. This decomposition uses basic trigonometry, allowing us to analyze each motion component separately.

Here's how it works:
  • **Horizontal Component (\( v_{0x} \))**: Since there are no horizontal forces (assuming no air resistance), this component remains constant throughout the flight. It is calculated by:\[ v_{0x} = v_0 \cos(\theta) \]where \( v_0 \) is the initial velocity and \( \theta \) is the angle above the horizontal.
  • **Vertical Component (\( v_{0y} \))**: This component changes over time due to the effect of gravity. It is initially calculated by:\[ v_{0y} = v_0 \sin(\theta) \]
Using the given conditions, we calculate these components as follows: the horizontal velocity \( v_{0x} \approx 24.0 \,\text{m/s} \) and the initial vertical velocity \( v_{0y} \approx 18.0 \,\text{m/s} \). Understanding these components helps in predicting the range and height of the projectile.
Vertical Motion Analysis
Vertical motion in projectile problems like this one involves examining how the component of velocity directed upwards or downwards changes due to gravity. It is crucial to discern how high the projectile goes and how long it remains in the air.

This is analyzed by:
  • Determining time intervals where specific vertical displacements occur, using the equation:\[ h = v_{0y} t - \frac{1}{2} g t^2 \]
  • Solving the quadratic equation derived from the above equation to find the times at which the projectile reaches a certain height, in our case, the 10 m mark.
  • Calculating the change in vertical velocity at those times, accounting for the effect of gravity, using:\[ v_y = v_{0y} - gt \]
In this exercise, by solving the quadratic equation, we found the baseball is at 10 meters at approximately 0.68 seconds and 3.0 seconds. By calculating the vertical velocities at these times, it's concluded that gravity reduces the vertical speed as it rises and subsequently increases it as it descends.

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Most popular questions from this chapter

When a train's velocity is 12.0 m/s eastward, raindrops that are falling vertically with respect to the earth make traces that are inclined 30.0\(^\circ\) to the vertical on the windows of the train. (a) What is the horizontal component of a drop's velocity with respect to the earth? With respect to the train? (b) What is the magnitude of the velocity of the raindrop with respect to the earth? With respect to the train?

In a World Cup soccer match, Juan is running due north toward the goal with a speed of 8.00 m/s relative to the ground. \(A\) teammate passes the ball to him. The ball has a speed of 12.0 m/s and is moving in a direction 37.0\(^\circ\) east of north, relative to the ground. What are the magnitude and direction of the ball's velocity relative to Juan?

The earth has a radius of 6380 km and turns around once on its axis in 24 h. (a) What is the radial acceleration of an object at the earth's equator? Give your answer in m/s\(^2\) and as a fraction of \(g\). (b) If \(a_{rad}\) at the equator is greater than \(g\), objects will fly off the earth's surface and into space. (We will see the reason for this in Chapter 5.) What would the period of the earth's rotation have to be for this to occur?

A baseball thrown at an angle of 60.0\(^{\circ}\) above the horizontal strikes a building 18.0 m away at a point 8.00 m above the point from which it is thrown. Ignore air resistance. (a) Find the magnitude of the ball's initial velocity (the velocity with which the ball is thrown). (b) Find the magnitude and direction of the velocity of the ball just before it strikes the building.

A faulty model rocket moves in the \(xy\)-plane (the positive \(y\)-direction is vertically upward). The rocket's acceleration has components \(a_x(t) = \alpha t^2\) and \(a_y(t) = \beta - \gamma t\), where \(\alpha = 2.50 m/s^4, \beta = 9.00 m/s^2,\) and \(\gamma = 1.40 m/s^3\). At \(t = 0\) the rocket is at the origin and has velocity \(\vec{v}_0=v_0\hat{i} + v_{0y}\hat{j}\) with \(v_{0x}\) = 1.00 m/s and \(v_{0y}\) = 7.00 m/s. (a) Calculate the velocity and position vectors as functions of time. (b) What is the maximum height reached by the rocket? (c) What is the horizontal displacement of the rocket when it returns to \(y = 0\)?
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