Problem 17 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

(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°.

Step 1: Decompose the Initial Velocity

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

Step 2: 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 \( t \): \[10 = 18t - 4.9t^2\]

Step 3: 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:

Step 4: 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.

Step 5: 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:\[v_y = 18.0 - 9.8 \times 3.0 \approx -11.4\,\text{m/s}\]

Step 6: 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 \( \theta \):\[\theta = \tan^{-1} \left(\frac{-18.0}{24.0}\right) \approx -36.9°\]This is below the horizontal.

Key Concepts

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.