Problem 50 - College Physics

Question

\(\bullet\) The longest home run. According to the Guinness Book of World Records, the longest home run ever measured was hit by Roy "Dizzy" Carlyle in a minor-league game. The ball traveled 188 \(\mathrm{m}(618 \mathrm{ft})\) before landing on the ground outside the ball- park. (a) Assuming that the ball's initial velocity was 45 above the horizontal, and ignoring air resistance, what did the initial speed of the ball need to be to produce such a home run if the ball was hit at a point 0.9 \(\mathrm{m}(3.0 \mathrm{ft})\) above ground level? Assume that the ground was perfectly flat. (b) How far would the ball be above a fence 3.0 \(\mathrm{m}(10 \mathrm{ft})\) in height if the fence were 116 \(\mathrm{m}\) \((380 \mathrm{ft})\) from home plate?

Step-by-Step Solution

Answer

(a) The initial speed needed is approximately 41.99 m/s. (b) The ball would be 13.4 m above the fence.

Step 1: Break Down the Problem

This problem involves projectile motion. There are two main parts: (a) finding the initial speed of the ball given a specific angle and conditions, and (b) determining the ball's height above a fence at a certain distance. We'll address each part separately.

Step 2: Initial Assumptions and Equations

Assume the angle of projection, \( \theta = 45^\circ \). We use the kinematic equations to solve the problem. The range formula for projectile motion is \( R = \frac{v_0^2 \sin(2\theta)}{g} \). Given that \( R = 188 \text{ m} \) and the initial height is 0.9 m, we incorporate these into our equations.

Step 3: Solving for Initial Speed (Part a)

1. Use the range formula for projectile motion adjusted for the initial height: \[ y = x \tan \theta - \frac{g x^2}{2 v_0^2 \cos^2 \theta} + 0.9 \] Set \( y = 0 \) when \( x = 188 \text{ m} \), rearrange to solve for the initial speed: \[ 0 = 188 \tan 45^\circ - \frac{9.8 \times 188^2}{2 v_0^2 \cos^2 45^\circ} + 0.9 \]2. Simplify: \[ v_0 = \sqrt{\frac{9.8 \times 188^2}{188 + 0.9}} \approx 41.99 \text{ m/s} \]

Step 4: Calculating Height Above Fence (Part b)

Using \( v_0 = 41.99 \text{ m/s} \) and \( \theta = 45^\circ \):1. Find the time \( t \) when \( x = 116 \text{ m} \): \[ x = v_0 \cos \theta \times t \Rightarrow t = \frac{116}{41.99 \cos 45^\circ} \approx 3.9 \text{ s} \]2. Use this time to find the vertical position: \[ y = v_0 \sin \theta \times t - \frac{1}{2}gt^2 + 0.9 \] Substitute values to find: \[ y = 41.99 \sin 45^\circ \times 3.9 - \frac{1}{2} \times 9.8 \times (3.9)^2 + 0.9 \approx 16.4 \text{ m} \]3. Determine height over the fence: \[ ext{Height over fence} = 16.4 - 3.0 = 13.4 \text{ m} \]

Key Concepts

Kinematic Equations

Projectile motion, like hitting a baseball, can be solved using kinematic equations. These equations are vital for calculating various aspects of a projectile's motion, such as displacement, velocity, and time. They allow us to analyze the path of an object under the influence of constant acceleration, typically due to gravity. In the context of the given problem, the kinematic equations help determine where and when the baseball lands, how fast it needs to travel to hit a certain distance, and its height at a given point in its trajectory.

Important equations used include:

• Vertical motion: \( y = v_0 \sin \theta \times t - \frac{1}{2}gt^2 + h_0 \)where \( y \) is the vertical position, \( v_0 \sin \theta \) represents the vertical component of the initial velocity, \( g \) is the acceleration due to gravity, \( t \) is the time, and \( h_0 \) is the initial height.
• Horizontal motion: \( x = v_0 \cos \theta \times t \)where \( x \) is the horizontal displacement, and \( v_0 \cos \theta \) is the horizontal component of the initial velocity.

These kinematic equations are derived from Newton's laws of motion, making them fundamental tools for analyzing projectile motion.

Range Formula

The range formula in projectile motion is used to find the horizontal distance a projectile will travel before touching the ground. It's particularly useful when you know the angle of projection and need to find how far an object, like a baseball, can be expected to travel.

The formula is: \[ R = \frac{v_0^2 \sin(2\theta)}{g} \]where:

• \( R \) is the range (horizontal distance).
• \( v_0 \) is the initial velocity of the projectile.
• \( \theta \) is the angle of projection.
• \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \) on Earth.

This formula is derived assuming a level launch and landing surface with no air resistance. It simplifies solving problems like the one in the exercise, helping predict the distance covered purely based on speed and angle.

Initial Velocity

Initial velocity is a crucial factor in projectile motion because it determines not only how far the object will travel but also how high it will go. It is the speed at which the projectile is launched and can be broken down into two components: horizontal and vertical. In the problem of the longest home run, the initial speed of the ball was calculated to propel it as far as required.

To separate these components, we use trigonometric functions:

• Horizontal component: \( v_{0x} = v_0 \cos \theta \)
• Vertical component: \( v_{0y} = v_0 \sin \theta \)

For a known angle of 45 degrees, these components are equal, simplifying calculations as the sine and cosine of 45 degrees are the same. This balance helps maximize the range, as 45 degrees is the optimal angle for distance in absence of other factors like air resistance.

Vertical Position

The vertical position of a projectile at any given time is crucial to understand its trajectory, altitude at specific points, and avoid obstacles during its flight. In our context, to determine how high above a fence the baseball would be, we calculated the vertical position at a certain horizontal distance from the launch point.

The equation to find vertical position considers several factors:\[ y = v_0 \sin \theta \times t - \frac{1}{2}gt^2 + h_0 \]Here, \( y \) represents the height above ground level, \( v_0 \sin \theta \) accounts for the initial upward velocity, the quadratically falling term accounts for gravity’s pull, and \( h_0 \) is the initial height from which it was hit. By solving this, you can find out how high the projectile is at any point in time, which is particularly useful when checking if it clears an obstacle like a fence.

Projectile Height

Estimating the maximum height a projectile reaches is just as important as knowing the range for understanding its motion. Projectile height not only relates to whether an object can clear obstacles but also provides insight into the energy and forces acting on it. For the home run problem, calculating the height above the fence checked if the ball's path was sufficient to pass over the stated obstacle.

The maximum height can be found by determining the point where the vertical velocity is zero (at the top of the trajectory). The kinetic energy and potential energy principles help derive the max height:\[ h_{ ext{max}} = \frac{v_{0y}^2}{2g} + h_0 \]where \( v_{0y} \) is the initial vertical velocity and \( h_0 \) is the launch height. Understanding projectile height is key for applied problems involving trajectory, like sports or physics predictions.