Problem 47 - Fundamentals of Physics

Question

A batter hits a pitched ball when the center of the ball is \(1.22 \mathrm{~m}\) above the ground. The ball leaves the bat at an angle of \(45^{\circ}\) with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of \(107 \mathrm{~m}\). (a) Does the ball clear a \(7.32\) -m-high fence that is \(97.5 \mathrm{~m}\) horizontally from the launch point? (b) At the fence, what is the distance between the fence top and the ball center?

Step-by-Step Solution

Answer

(a) The ball clears the fence. (b) The distance is 16.5 m.

Step 1: Understand the Problem

We need to determine if a baseball, hit with specific initial conditions, clears a 7.32 m high fence located 97.5 m away horizontally. The ball starts at 1.22 m above the ground and is hit at a 45° angle. We are given the range as 107 m if it returns to the launch level.

Step 2: Use Projectile Motion Equations

The key formula for the projectile's height as a function of horizontal distance is:\[ y(x) = x \tan(\theta) - \left( \frac{g}{2v_0^2 \cos^2(\theta)} \right) x^2 + h_0 \]where \( \theta = 45^{\circ} \), \( g \approx 9.81 \ \text{m/s}^2 \), \( x \) is horizontal distance, \( y \) is height, \( v_0 \) is initial velocity, and \( h_0 = 1.22 \ \text{m} \). This formula can find the height of the ball at any given horizontal distance \( x \).

Step 3: Calculate Initial Velocity

Since the range is given by:\[ R = \frac{v_0^2 \sin(2\theta)}{g} \]plugging in the known values, we have:\[ 107 = \frac{v_0^2 \sin(90^{\circ})}{9.81} \]Solving for \( v_0 \), we find:\[ v_0 = \sqrt{107 \times 9.81} \approx 32.5 \ \text{m/s} \]

Step 4: Find the Height at the Fence Location

Using the projectile motion equation and the calculated \( v_0 \), find the height \( y \) at \( x = 97.5 \ \text{m} \):\[ y(97.5) = 97.5 \tan(45^{\circ}) - \left( \frac{9.81}{2 \cdot (32.5)^2 \cdot \cos^2(45^{\circ})} \right) (97.5)^2 + 1.22 \]Simplify to get:\[ y(97.5) \approx 23.82 \ \text{m} \]

Step 5: Compare Height with the Fence

The ball's height at 97.5 m is approximately 23.82 m, while the fence height is 7.32 m. Since 23.82 m is greater than 7.32 m, the ball clears the fence.

Step 6: Calculate the Distance Between Ball and Fence Top

Subtract the fence height from the ball's height at the fence location: \[ 23.82 - 7.32 = 16.5 \ \text{m} \]Thus, the distance between the fence top and ball center is 16.5 m.

Key Concepts

Initial Velocity

Initial velocity is the speed at which an object starts its motion. It is a crucial factor in determining how a projectile travels through the air. For the baseball scenario, initial velocity is the speed at which the ball leaves the bat. It plays a significant role in both the range and the maximum height of the projectile.

The formula to calculate initial velocity, given the range, is:\[R = \frac{v_0^2 \sin(2\theta)}{g}\]Where:

• \( R \) is the range, which is the horizontal distance the ball covers.
• \( v_0 \) is the initial velocity.
• \( \theta \) is the launch angle.
• \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).

For this problem, by plugging in the given range (107 m) and solving for \( v_0 \), the initial velocity is calculated to be approximately 32.5 m/s. This speed ensures that the projectile has the necessary energy to achieve the specified horizontal range.

Horizontal Distance

Horizontal distance, or range, refers to how far a projectile travels parallel to the ground. In projectile motion, horizontal distance is influenced by the initial velocity and the angle at which the projectile is launched.

Here, the task is to understand whether the baseball, hit with a range capacity of 107 m, can clear a fence placed at 97.5 m. The range of 107 m indicates the distance the projectile will cover when it lands or reaches the same vertical level from which it was launched. Thus, analyzing whether the ball at any point exceeds the fence’s height helped determine its clearance over the fence.
This horizontal distance forms a crucial aspect of the projectile motion, allowing us to predict important outcomes such as the exact landing spot and trajectory features.

Launch Angle

The launch angle is the angle at which an object is projected into the air. It significantly affects the trajectory and range of the projectile. In the given problem, the launch angle is 45 degrees.

A 45-degree launch angle is often regarded as the optimal angle for maximizing range in projectile motion. This is because it allows the initial velocity to be divided equally between the horizontal and vertical components.

• The horizontal component determines how far the projectile will travel.
• The vertical component determines how high the projectile will go.

By using this launch angle, the baseball not only efficiently utilizes its initial velocity but also ensures an effective balance between distance and height. This angle is particularly important because it influences how far and how high the ball will travel before gravity pulls it back to the ground.

Projectile Height

Projectile height refers to the altitude a projectile reaches during its flight. In this context, it is the maximum height the baseball attains above the ground level. It's lifted by the initial velocity and launch angle, opposed by gravitational forces.

Using the projectile motion formula, the height can be determined at any point along the projectile's path based on its horizontal distance. For instance, at 97.5 m, the height calculated is approximately 23.82 m, which is significantly greater than the fence's height of 7.32 m, indicating the ball comfortably clears the fence.
This height computation involves accounting for the initial vertical position, launch angle, and velocity to consider both upward lift and downward gravitational pull, depicting how the projectile moves above its launching pad.