Problem 87 - Fundamentals of Physics

Question

A baseball is hit at ground level. The ball reaches its maximum height above ground level \(3.0 \mathrm{~s}\) after being hit. Then \(2.5 \mathrm{~s}\) after reaching its maximum height, the ball barely clears a fence that is \(97.5 \mathrm{~m}\) from where it was hit. Assume the ground is level. (a) What maximum height above ground level is reached by the ball? (b) How high is the fence? (c) How far beyond the fence does the ball strike the ground?

Step-by-Step Solution

Answer

(a) 44.1 m (b) 13.475 m (c) 53.18 m beyond the fence.

Step 1: Analyze Part (a) - Understand Maximum Height

The ball reaches its maximum height 3.0 seconds after being hit. At the maximum height, the velocity in the vertical direction becomes zero. The time it takes for a projectile to reach its maximum height can be used to find the maximum height using the formula for displacement in uniformly accelerated motion:\[h = v_0 t - \frac{1}{2} g t^2\]where \(v_0\) is the initial velocity in the vertical direction, \(t = 3.0\) s is the time to reach maximum height, and \(g = 9.8\, \text{m/s}^2\) is the acceleration due to gravity.

Step 2: Calculate Initial Vertical Velocity

We rearrange the kinematic equation for the velocity to solve for the initial vertical velocity at maximum height, knowing the final velocity at this point is zero:\[v = v_0 - gt\]Set \(v = 0\):\[0 = v_0 - g \times 3.0\]Solving for \(v_0\) gives:\[v_0 = g \times 3.0 = 9.8 \times 3 = 29.4 \text{ m/s}\]

Step 3: Calculate Maximum Height

Substitute the initial vertical velocity back into the displacement formula to find the maximum height:\[h = 29.4 \times 3 - \frac{1}{2} \times 9.8 \times (3)^2\]Calculate:\[h = 88.2 - 44.1 = 44.1 \text{ meters}\]

Step 4: Analyze Part (b) - Height of the Fence

The problem states that the ball barely clears the fence 2.5 seconds after reaching the maximum height. The horizontal distance to the fence is 97.5 meters. To find the vertical position of the ball, we calculate its position 2.5 seconds after it starts to fall from the maximum height using:\[y = h - \frac{1}{2} g t^2\]where \(t = 2.5\) s.

Step 5: Calculate Vertical Position at Fence

Using the expression for vertical displacement during the descent:\[y = 44.1 - \frac{1}{2} \times 9.8 \times (2.5)^2\]Calculate:\[y = 44.1 - 30.625 = 13.475 \text{ meters}\]

Step 6: Analyze Part (c) - Ground Strike Beyond the Fence

The total time from launch until returning to ground level can be obtained by doubling the time it takes to reach maximum height plus time to fall an additional 2.5 seconds to the fence:\[total\_time = 2 \times 3.0 + 2.5 = 8.5 \text{ seconds}\]We need to calculate how much farther the ball travels beyond the fence horizontally. The initial horizontal velocity \(v_{0x}\) can be calculated using the horizontal distance to the fence.

Step 7: Calculate Horizontal Velocity

The horizontal distance to the fence is 97.5 meters, which is covered in 5.5 seconds (3.0 to hit the maximum height + 2.5 to descend to the fence). Using\[v_{0x} = \frac{97.5}{5.5}\]Calculate:\[v_{0x} = 17.727 \text{ m/s}\]

Step 8: Total Horizontal Distance Calculation

Calculate the total distance the ball travels horizontally using the total time until ground-strike:\[d = v_{0x} \times total\_time\]Substitute the horizontal velocity and total time:\[d = 17.727 \times 8.5 = 150.68\text{ meters}\]

Step 9: Calculate Distance Beyond the Fence

Subtract the distance to the fence from the total horizontal distance traveled:\[d_{beyond} = 150.68 - 97.5 = 53.18 \text{ meters}\]

Key Concepts

Kinematics

Kinematics is the branch of physics that deals with understanding the motion of objects. When examining projectile motion, kinematics allows us to analyze an object's trajectory without considering the forces causing the motion. Kinematics divides motion into different types, and one of the main focus areas is understanding how objects move under gravity. This forms the basis of analyzing projectile motion, where the vertical and horizontal motions are examined separately.
By identifying the initial conditions such as initial velocity and angle of projection, we can predict the future position of the projectile using kinematic equations. Key concepts include calculating the time it takes for an object to reach peak height, the maximum height achieved, and the distance it will travel horizontally before hitting the ground. The use of these equations allows us to understand the entire trajectory of the projectile.

Uniformly Accelerated Motion

Uniformly accelerated motion refers to the movement where an object’s acceleration remains constant over time. In the context of projectile motion, the most common and influential constant acceleration is due to gravity, which acts downward with a magnitude of approximately 9.8 m/s². In the vertical dimension, projectile motion is a classic example of uniformly accelerated motion, as gravity consistently pulls the object towards the ground. This leads to a change in the velocity of the projectile, affecting how far it goes, how long it stays in the air, and the height it reaches. For example, when calculating the maximum height of a projectile or determining how long it will take to return to the ground, you can apply the formulas of uniformly accelerated motion, such as:

• Final velocity (v) = Initial velocity (v₀) + acceleration (a) × time (t)
• Displacement (s) = Initial velocity (v₀) × time (t) + ½ × acceleration (a) × time² (t²)

These equations are essential tools for solving problems involving projectiles.

Vertical and Horizontal Components of Motion

In projectile motion, it's crucial to separate the motion into vertical and horizontal components. These components operate independently of each other, with the only common factor being time.

Vertical Motion

For the vertical component, gravity is the sole force acting on the projectile, causing it to decelerate as it ascends and accelerate as it descends. Calculating the vertical speed at any given moment involves considering gravity's impact and using kinematic equations for uniformly accelerated motion.

Horizontal Motion

In contrast, the horizontal component of motion generally experiences no acceleration (assuming air resistance is negligible), meaning the horizontal velocity remains constant. The distance covered horizontally can be calculated using the formula:\[ x = v_{0x} \times t\]which is applicable when the velocity is consistent, noting that \(x\) is the horizontal displacement, \(v_{0x}\) is the initial horizontal velocity, and \(t\) represents time.By isolating these two components, we simplify the analysis of projectile motion, allowing for more straightforward calculations of different parameters such as range, flight time, and maximum height.

Physics Problems

Physics problems involving projectile motion challenges often require calculating multiple variables and understanding different motion components. These problems test understanding of both the formulas and the underlying principles of motion.Approach problems by first identifying knowns and unknowns, then applying relevant physics principles and equations to solve them. For instance:

• Identify initial conditions: initial speed, angle of projection, height of release
• Calculate vertical parameters: time to peak height using \( v = v_0 - gt \) and maximum height using \( h = v_0 t - \frac{1}{2} g t^2 \)
• Find horizontal values: reach the to-fence distance using constant speed formulas

Real-world applications and typical problem-solving include predicting the projectile's behavior when encountering obstacles or determining precise landing spots based on variable changes. Working systematically through each step ensures clarity and accuracy in solving complex narratives of physics problems.