Problem 38 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A baseball is hit with a speed of 27.0 \(\mathrm{m} / \mathrm{s}\) at an angle of \(45.0^{\circ} .\) It lands on the flat roof of a \(13.0-\mathrm{m}\) -tall nearby building. If the ball was hit when it was 1.0 \(\mathrm{m}\) above the ground, what horizontal distance does it travel before it lands on the building?

Step-by-Step Solution

Answer

The baseball travels approximately 59.2 meters horizontally.

Step 1: Break Down the Problem

Identify the known variables and the requirements. We know the initial speed \( v_0 = 27.0 \mathrm{m/s} \), launch angle \( \theta = 45.0^{\circ} \), height of the building \( h = 13.0 \mathrm{m} \), and initial height \( h_0 = 1.0 \mathrm{m} \). We aim to find the horizontal distance the ball travels before landing.

Step 2: Resolve Initial Velocity Components

Use the initial speed and angle to find the horizontal and vertical components of the velocity. The formulas are:\[v_{0x} = v_0 \cdot \cos(\theta)\]\[v_{0y} = v_0 \cdot \sin(\theta)\]Substituting the known values:\[v_{0x} = 27.0 \cdot \cos(45.0^{\circ}) \approx 19.1 \mathrm{m/s}\]\[v_{0y} = 27.0 \cdot \sin(45.0^{\circ}) \approx 19.1 \mathrm{m/s}\]

Step 3: Use Kinematic Equation for Vertical Motion

Use the vertical motion equation to find the time of flight until the ball reaches 13.0 m. The equation is:\[y = h_0 + v_{0y}t - \frac{1}{2}gt^2\]Where \( y = 13.0 \mathrm{m} \) and \( g = 9.8 \mathrm{m/s^2} \). Plugging in the values:\[13 = 1.0 + 19.1t - 4.9t^2\]Simplify to:\[4.9t^2 - 19.1t + 12 = 0\]

Step 4: Solve the Quadratic Equation

Use the quadratic formula to solve for \( t \):\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Where \( a = 4.9 \), \( b = -19.1 \), \( c = 12 \). Calculating the values:\[t = \frac{19.1 \pm \sqrt{(-19.1)^2 - 4 \cdot 4.9 \cdot 12}}{9.8}\]\[t = \frac{19.1 \pm \sqrt{364.81 - 235.2}}{9.8} \]\[t = \frac{19.1 \pm \sqrt{129.61}}{9.8} \]\[t = \frac{19.1 \pm 11.38}{9.8}\]The positive root is \( t \approx 3.10 \mathrm{s} \).

Step 5: Find Horizontal Distance

Calculate the horizontal distance \( x \) using the horizontal velocity and time:\[x = v_{0x} \cdot t\]Plugging in the values:\[x = 19.1 \cdot 3.10 \approx 59.2 \mathrm{m}\]

Step 6: Conclusion

The baseball travels approximately 59.2 meters horizontally before landing on the building.

Key Concepts

Initial Velocity Components

Understanding the initial velocity components in projectile motion is crucial for solving the problem of a baseball hit onto a building rooftop. When a projectile is launched at an angle, its velocity can be divided into two components: horizontal and vertical. These components help determine how far and how high the projectile will travel.

• **Horizontal Velocity (**\(v_{0x}\)): This component affects how fast the projectile travels along the horizontal plane. It is calculated using the formula \(v_{0x} = v_0 \cdot \cos(\theta)\), where \(v_0\) is the initial speed and \(\theta\) is the launch angle.
• **Vertical Velocity (**\(v_{0y}\)): This component dictates how the projectile moves vertically and is found using \(v_{0y} = v_0 \cdot \sin(\theta)\). It affects both the time the projectile stays in the air and how high it will rise.

This breakdown into components simplifies the mathematical handling of projectile motion, allowing use of straightforward kinematic equations for each direction.

Kinematic Equations

Kinematic equations describe the motion of a projectile under the influence of gravity. They are pivotal in determining parameters like time of flight, maximum height, and range of the projectile.When dealing with vertical motion, we use the equation:\[y = h_0 + v_{0y}t - \frac{1}{2}gt^2\]Here, \(y\) represents the final vertical position, \(h_0\) is the initial vertical position, \(v_{0y}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity \((9.8 \, m/s^2)\), and \(t\) is the time. This formula incorporates the effects of gravity, making it possible to solve for the time at which the projectile reaches a specific height.Since we also need to consider horizontal motion, the horizontal distance \(x\) can be found from:\[x = v_{0x} \cdot t\]This equation indicates that horizontal motion assumes constant velocity because gravity affects it perpendicularly, not directly.

Quadratic Formula

The quadratic formula is a mathematical tool used to find solutions to quadratic equations, which is essential when determining the time of flight in non-linear motion such as projectile motion.Given a quadratic equation in the standard form:\[ax^2 + bx + c = 0\]The solutions for \(x\) are given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In our case, the quadratic equation, derived from the kinematic equation, relates to time \(t\) for a projectile's vertical motion. The coefficients correspond to:

• \(a\) as the coefficient for the squared term, representing \( \frac{1}{2}g \).
• \(b\) as the linear coefficient, representing the initial vertical velocity \(v_{0y}\).
• \(c\) as the constant term, derived from initial and desired heights.

Simplifying these calculation steps with the quadratic formula helps find the exact time when the projectile will reach the target height, which is crucial for later computing the horizontal distance traveled.