Question

Bill Jones has a bad night in his bowling league. When he gets home, he drops his bowling ball in disgust out the window of his apartment, from a height of \(63.17 \mathrm{~m}\) above the ground. John Smith sees the bowling ball pass by his window when it is \(40.95 \mathrm{~m}\) above the ground. How much time passes from the time when John Smith sees the bowling ball pass his window to when it hits the ground?

Short answer

Answer: It takes about 2.01 seconds for the bowling ball to reach the ground from the height of John Smith's window.

Step-by-step solution

Find the initial velocity when the ball passes John's window at 40.95 m

We will use the following free-fall formula to determine the initial velocity (\(v_0\)) when the ball passes John Smith's window at a height of 40.95 m: \[ v^2 = v_0^2 + 2as \] where \(v\) is the final velocity, \(a\) is the acceleration due to gravity (\(-9.81 \mathrm{~m/s^2}\)), and \(s\) is the distance fallen (displacement) from the initial height. With the initial height of 63.17 m above the ground and the height of John's window at 40.95 m, we have: \[ s = 63.17 - 40.95 = 22.22 \mathrm{~m} \] Now, let's rewrite the free-fall formula as: \[ v_0^2 = v^2 - 2as \] We know that the final velocity (\(v\)) is 0 m/s when the ball reaches John's window. So, we have: \[ v_0^2 = 0^2 - 2(-9.81)(22.22) \] Calculating \(v_0\), we get: \[ v_0 = \sqrt{(2)(9.81)(22.22)} = 20.82 \mathrm{~m/s} \]

Calculate the time for the ball to reach the ground

Now that we have the initial velocity at the height of 40.95 m, we can use the following equation to find the time (\(t\)) it takes for the ball to reach the ground: \[ s = v_0t + \frac{1}{2}at^2 \] In this case, the height fallen (\(s\)) is 40.95 m. We also know that the acceleration due to gravity (\(a\)) is -9.81 m/s² and the initial velocity (\(v_0\)) is 20.82 m/s. Plugging these values into the equation, we get: \[ 40.95 = 20.82t + \frac{1}{2}(-9.81)t^2 \] To solve for \(t\), we can rewrite the equation as a quadratic equation: \[ 4.905t^2 - 20.82t + 40.95 = 0 \] Now, solve the quadratic equation. There are two possible solutions, but we choose the positive one as time cannot be negative. We get: \[ t = 2.01 \mathrm{~seconds} \] So, it takes about 2.01 seconds for the bowling ball to reach the ground from the height of John Smith's window.

Key concepts

Understanding Kinematics in Free-Fall Motion

Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. In the context of free-fall motion, kinematics allows us to describe the motion of an object dropped from a certain height, such as a bowling ball falling from a window.

When studying free-fall motion, we make a few key assumptions: firstly, we ignore air resistance, and secondly, we consider the acceleration due to gravity to be constant, at approximately \( -9.81 \text{ m/s}^2 \) downward. Using kinematic equations, we can calculate various parameters of the motion, including velocity, displacement, and time.

The kinematic equation we used to solve for time in the given exercise was

\( s = v_0t + \frac{1}{2}at^2 \)

where \(s\) is the displacement, \(v_0\) is the initial velocity, \(t\) is the time, and \(a\) is the acceleration due to gravity. When the bowling ball drops and passes John Smith's window, these variables allow us to calculate how long it takes for the ball to hit the ground from that point.

The Role of Quadratic Equations in Motion Analysis

Quadratic equations are polynomial equations of the second degree, typically taking the form

\( ax^2 + bx + c = 0 \)

In motion problems involving constant acceleration, such as the free-fall example with the bowling ball, the kinematic equations can be rearranged into quadratic form. This is because the motion equation for displacement with constant acceleration includes a term with time squared.

By setting up the kinematic equation in the quadratic form, as seen in the exercise, we can solve for time using quadratic solution methods. In the case at hand,

\( 4.905t^2 - 20.82t + 40.95 = 0 \)

is the quadratic equation we derived from the kinematic equation. To find the time \(t\), we must solve for the roots. We disregard the negative root because time cannot be negative and accept the positive root as the physical solution. Quadratic equations are a fundamental tool in understanding the dynamics of free-fall motion and enable us to calculate precise outcomes such as the time it takes for an object to hit the ground.

Acceleration Due to Gravity: A Constant Force in Free-Fall

Acceleration due to gravity is a vital concept in understanding free-fall motion. On Earth's surface, this acceleration is constant and has a value of \(9.81 \text{ m/s}^2 \) directed downwards; it is denoted by the symbol \(g\). This constant acceleration influences every object in free-fall, imparting the same rate of increase in speed, irrespective of the object's mass. The negative sign \( -9.81 \text{ m/s}^2 \) used in the equations reflects the direction of the acceleration, which is towards the center of the Earth.

In kinematic equations, \(g\) replaces the general acceleration \(a\), and it's used to predict the motion of an object under free-fall. For instance, the bowling ball in the example experiences this acceleration, and its velocity increases uniformly as it falls. The acceleration due to gravity is crucial for calculating both the initial velocity of the ball as it passes John's window and the time it takes to reach the ground. It is also a universal force experienced by all objects in free-fall near the Earth's surface, making it a cornerstone concept for many physics problems.