Problem 37 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A baseball is hit with a specd 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 horizontal distance is approximately 59.4 meters.

Step 1: Determine Initial Horizontal and Vertical Velocities

Using the initial speed of the baseball, 27.0 m/s, we can resolve it into horizontal and vertical components using the angle of projection: \[ v_{0x} = v_0 \cos \theta = 27.0 \, \text{m/s} \times \cos 45^{\circ} \]\[ v_{0y} = v_0 \sin \theta = 27.0 \, \text{m/s} \times \sin 45^{\circ} \]Since \( \cos 45^{\circ} = \sin 45^{\circ} = \frac{\sqrt{2}}{2} \):\[ v_{0x} = v_{0y} = 27.0 \times \frac{\sqrt{2}}{2} \approx 19.1 \, \text{m/s} \]

Step 2: Calculate the Time of Flight

The time the ball is in the air is determined by examining the vertical motion. The equation for vertical motion is:\[ y = y_0 + v_{0y} t - \frac{1}{2} g t^2 \]Here, \(y_0 = 1.0\) m, \(y = 13.0\) m (final height), \(g = 9.8 \, \text{m/s}^2\):\[ 13.0 = 1.0 + 19.1t - 4.9t^2 \]\[ 4.9t^2 - 19.1t + 12 = 0 \]Solving this quadratic equation using the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a = 4.9\), \(b = -19.1\), \(c = 12\):\[ t = \frac{19.1 \pm \sqrt{(19.1)^2 - 4 \times 4.9 \times 12}}{2 \times 4.9} \]Calculate the solutions, taking the positive root since time cannot be negative.

Step 3: Quadratic Formula Calculation

Calculate the discriminant:\[ (19.1)^2 - 4 \times 4.9 \times 12 = 364.81 - 235.2 = 129.61 \]Now substitute:\[ t = \frac{19.1 \pm \sqrt{129.61}}{9.8} \]\[ t = \frac{19.1 \pm 11.38}{9.8} \]We choose the positive solution:\[ t \approx \frac{19.1 + 11.38}{9.8} \approx \frac{30.48}{9.8} \approx 3.11 \, \text{seconds} \]

Step 4: Calculate Horizontal Distance

Using the horizontal velocity component and the time of flight, calculate the horizontal distance:\[ d = v_{0x} \times t \]\[ d = 19.1 \, \text{m/s} \times 3.11 \, \text{s} \approx 59.4 \, \text{m} \]

Key Concepts

Initial Velocity Components

When a projectile is launched, it has an initial speed and angle of launch that determine how it will move through the air. To analyze this movement, we break the initial speed into two parts: horizontal and vertical components.

• The horizontal component \(v_{0x}\) is found using the cosine of the launch angle: \(v_{0x} = v_0 \cos \theta\).
• The vertical component \(v_{0y}\) is found using the sine of the launch angle: \(v_{0y} = v_0 \sin \theta\).

In this exercise, with an initial speed of 27.0 m/s and an angle of 45°, both components are equal because \(\cos 45^{\circ}\) and \(\sin 45^{\circ}\) are the same, each about 0.707. This results in: \(v_{0x} = v_{0y} \approx 19.1 \text{ m/s}\). This means the baseball will travel equally fast horizontally and vertically at the start of its flight.

Time of Flight

The time of flight refers to how long a projectile is in the air. For vertical motion, the equation used is:\[y = y_0 + v_{0y} t - \frac{1}{2} g t^2\]where:

• \(y\) is the final vertical position,
• \(y_0\) is the initial vertical position,
• \(v_{0y}\) is the initial vertical velocity,
• \(g\) is the acceleration due to gravity (9.8 m/s²),
• \(t\) is the time of flight.

The ultimate goal is to solve for \(t\) when the ball reaches the building's roof at 13.0 m, having started from a height of 1 m. The equation becomes a quadratic in the form \(4.9t^2 - 19.1t + 12 = 0\), which can be solved using the quadratic formula.

Quadratic Equation in Physics

The quadratic equation often appears in physics problems, especially in projectile motion scenarios. With parameters like initial speed and height, it helps us find key times or distances.\[at^2 + bt + c = 0\]where \(a\), \(b\), and \(c\) are coefficients. In our exercise, they were determined to be:

• \(a = 4.9\)
• \(b = -19.1\)
• \(c = 12 \)

The quadratic formula gives the time \(t\):\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]By calculating the discriminant \((b^2 - 4ac)\), one finds whether real solutions exist, which then gives valid times. From the solutions, the physically meaningful positive root \(t \approx 3.11 \text{ seconds}\) is chosen, representing the ball's time to reach the building's roof. This helps in further calculations like determining how far the baseball travels horizontally.