Problem 61 - Fundamentals of Physics

Question

A steel ball is dropped from a building's roof and passes a window, taking \(0.125 \mathrm{~s}\) to fall from the top to the bottom of the window, a distance of \(1.20 \mathrm{~m}\). It then falls to a sidewalk and bounces back past the window, moving from bottom to top in \(0.125 \mathrm{~s}\). Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is \(2.00 \mathrm{~s}\). How tall is the building?

Step-by-Step Solution

Answer

The building is 24.38 meters tall.

Step 1: Calculate the velocity of the ball

To calculate the velocity of the ball as it passes by the window, use the equation for constant acceleration: \( v = \frac{d}{t} \), where \( d = 1.20 \text{ m} \) and \( t = 0.125 \text{ s} \). Therefore, \( v = \frac{1.20}{0.125} = 9.6 \text{ m/s} \). This is the average velocity as it passes the window since gravity is acting on the ball.

Step 2: Determine the velocity just below the window

Since the average velocity over the window is \( 9.6 \text{ m/s} \) and the fall is symmetrical, the velocity just before and after the window is the same \( v = 9.6 \text{ m/s} \).

Step 3: Find the initial velocity of the ball using the window

Using the equation \( v = u + at \), where \( v = 9.6 \text{ m/s} \), \( a = 9.8 \text{ m/s}^2 \), and \( t = 0.125 \text{ s} \), solve for \( u \), the velocity of the ball just before it passed the window: \[ 9.6 = u + 9.8 \cdot 0.125 \].Thus, \( u = 9.6 - 1.225 = 8.375 \text{ m/s} \).

Step 4: Calculate the height of the building above the window

Using the kinematic equation \( v^2 = u^2 + 2as \), where \( v = 8.375 \text{ m/s} \), \( u = 0 \text{ m/s} \) (the velocity at the roof), and \( a = 9.8 \text{ m/s}^2 \), solve for \( s \):\[ (8.375)^2 = 0 + 2 \times 9.8 \times s \]\[ s = \frac{70.156}{19.6} = 3.58 \text{ m} \].

Step 5: Calculate the total height of the building

The total height of the building is the sum of the height above the window, the height of the window itself, and the height below the window. The ball spends \(2.00 \text{ s}\) below the window, and using \( s = ut + \frac{1}{2}at^2 \) with \( u = 0 \text{ m/s} \) (since it comes to rest) and \( a = 9.8 \text{ m/s}^2 \), we find:\[ s = 0 + \frac{1}{2} \times 9.8 \times (2.00)^2 = 19.6 \text{ m} \].Adding all distances: \( 3.58 + 1.20 + 19.6 = 24.38 \text{ m} \).

Key Concepts

Free Fall

Free fall is a simple yet fascinating physical concept where an object moves under the influence of gravitational force only. Other forces, such as air resistance, are typically ignored for simplicity unless specified otherwise. In free fall, an object accelerates downward due to Earth's gravity. This acceleration is constant at approximately 9.8 meters per second squared (m/s²).

This means that for each second an object is in free fall, its velocity increases by 9.8 m/s, starting from rest. Free fall can be observed in various scenarios, such as a steel ball being dropped from a building's roof. Although it seems straightforward, understanding free fall is crucial for solving problems related to vertical motion, like determining how long an object takes to hit the ground.

Constant Acceleration

Constant acceleration, or uniform acceleration, occurs when an object's velocity changes at a steady rate over time. In kinematics, constant acceleration scenarios are common, especially when analyzing the motion of falling objects like a steel ball dropped from a height.

Gravity provides a constant acceleration of 9.8 m/s² to freely falling bodies. This steady rate of change in velocity allows us to apply kinematic equations to calculate distances, velocities, and times without the complexity of varying acceleration rates. Understanding constant acceleration is key to solving problems involving free-falling objects and determining how they move over time.

Velocity Calculation

Velocity calculation is an essential part of kinematics and involves determining how fast an object is moving in a particular direction. When analyzing free-fall situations, knowing how to calculate velocity at different moments of the fall helps us understand the object's motion more comprehensively.

For instance, if a steel ball falls past a window with a known time and distance, we can calculate the average velocity using the formula \( v = \frac{d}{t} \), where \( d \) is the distance traveled, and \( t \) is the time taken. This method provides insights into how the ball's speed changes as it accelerates due to gravity. Regularly calculating velocities is critical for understanding kinematic problems and predicting motion outcomes.

Kinematic Equations

Kinematic equations are powerful tools in physics that describe the motion of objects with constant acceleration. These equations link displacement, initial velocity, final velocity, time, and acceleration, allowing us to solve various kinematic problems efficiently.

The core kinematic equations include:

• \( v = u + at \) — Final velocity \( v \) given initial velocity \( u \), acceleration \( a \), and time \( t \).
• \( s = ut + \frac{1}{2}at^2 \) — Displacement \( s \) given initial velocity, acceleration, and time.
• \( v^2 = u^2 + 2as \) — Final velocity given initial velocity, acceleration, and displacement.

These equations are crucial when solving problems involving vertical motion, such as calculating the height of a building from which a steel ball is dropped. Mastering the use of kinematic equations is essential for students to accurately predict how objects move.