Question

On August 24,1894 , Pop Shriver of the Chicago White Stockings caught a baseball dropped (by Clark Griffith) from the top of the Washington Monument. The Washington Monument is \(555 \mathrm{ft}\) tall and a baseball weighs \(5 \frac{1}{8} \mathrm{oz}\). (a) If we ignore air resistance and assume the baseball was acted upon only by gravity, how fast would the baseball have been traveling when it was \(7 \mathrm{ft}\) above the ground? (b) Suppose we now include air resistance in our model, assuming that the drag force is proportional to velocity with a drag coefficient \(k=0.0018 \mathrm{lb}-\mathrm{sec} / \mathrm{ft}\). How fast is the baseball traveling in this case when it is \(7 \mathrm{ft}\) above the ground?

Short answer

Answer: Without air resistance, the velocity of the baseball when it is 7 feet above the ground is approximately 144.6 ft/s. With air resistance, the velocity is approximately 106.87 ft/s.

Step-by-step solution

Calculate the free-fall distance

We first need to determine the height the baseball fell from. The Washington Monument is \(555\mathrm{ft}\) tall, and we want to know the velocity of the baseball when it's \(7\mathrm{ft}\) above the ground. So the total distance the baseball fell is: \(555\mathrm{ft} - 7\mathrm{ft} = 548\mathrm{ft}\).

Determine the velocity under free-fall condition

For part (a), we'll find the velocity of the baseball \(7\mathrm{ft}\) above the ground without considering air resistance. For this, we use the free-fall equation: \(v^2 = v_0^2 + 2gh\) where \(v\) is the final velocity, \(v_0\) is the initial velocity, \(g\) is the acceleration due to gravity (\(32\mathrm{ft/s^2}\)), and \(h\) is the distance fallen. Since the baseball was dropped, its initial velocity (\(v_0\)) is \(0\). Plug the other values into the equation: \(v^2 = 0^2 + 2(32\mathrm{ft/s^2})(548\mathrm{ft})\) Solve for \(v\) to find the velocity of the baseball \(7\mathrm{ft}\) above the ground without considering air resistance: \(v = \sqrt{2(32\mathrm{ft/s^2})(548\mathrm{ft})} \approx 144.6\mathrm{ft/s}\)

Determine the velocity including air resistance

For part (b), we'll find the velocity of the baseball \(7\mathrm{ft}\) above the ground considering air resistance. For this, we use the equation for terminal velocity in the presence of air resistance: \(v=\sqrt{\frac{2mg}{k}}\) where \(v\) is the terminal velocity, \(m\) is the mass of the baseball, \(g\) is the acceleration due to gravity (\(32\mathrm{ft/s^2}\)), and \(k\) is the drag coefficient. First, convert the weight of the baseball from ounces to pounds by dividing by \(16\mathrm{oz}\): \(5\frac{1}{8}\mathrm{oz} = 5.125\mathrm{oz} \cdot \frac{1\mathrm{lb}}{16\mathrm{oz}} \approx 0.3203\mathrm{lb}\) Now, plug these values into the equation for terminal velocity: \(v=\sqrt{\frac{2(0.3203\mathrm{lb})(32\mathrm{ft/s^2})}{0.0018\mathrm{lb\cdot sec/ft}}}\) Solve for \(v\) to find the velocity of the baseball \(7\mathrm{ft}\) above the ground considering air resistance: \(v \approx 106.87\mathrm{ft/s}\) So, the velocity of the baseball when it's \(7\mathrm{ft}\) above the ground without air resistance is approximately \(144.6\mathrm{ft/s}\), and with air resistance, it is approximately \(106.87\mathrm{ft/s}\).

Key concepts

Free-Fall Motion

Free-fall motion refers to the movement of an object where only the force acting on it is gravity. In this context, we look at how objects accelerate as they fall. When Pop Shriver caught a baseball dropped from the Washington Monument, it was a classic example of free-fall motion. For a falling object, like our baseball, gravity pulls it towards the Earth. Yet, without other forces to counteract it, the ball will continue to accelerate.When we calculate the speed of the baseball just before it hits the ground, we use the free-fall equation: \[ v^2 = v_0^2 + 2gh \]- \( v \) represents final velocity.- \( v_0 \) is the initial velocity, which is zero since it's dropped, not thrown.- \( g \) is acceleration due to gravity, around \( 32 \text{ft/s}^2 \) for our scenario.- \( h \) is the distance the object falls.Thus, free-fall gives us a simple way to measure how fast an object will be moving just before it reaches a certain point, like 7 feet above the ground in our example.

Air Resistance

When an object like a baseball falls through the air, it encounters a force known as air resistance. This force opposes the object's descent, acting in the direction opposite to the motion. Air resistance can significantly affect the object's speed as it falls.Air resistance is crucial for real-life modeling because:

• It reduces the speed of fast-moving objects.
• It depends on factors like speed, surface area, and air density.
• It can be modeled as a force proportional to the object's velocity.

For our baseball scenario, a drag coefficient \( k = 0.0018 \text{lb-sec/ft} \) is used to illustrate air resistance. This coefficient helps us calculate the effect of air on the baseball's velocity, showing that the drag force effectively slows down the baseball compared to an idealized free-fall scenario. Therefore, including air resistance paints a more accurate picture of the ball's descent.

Terminal Velocity

Terminal velocity is the highest speed an object can achieve as it falls through a fluid, such as air. It occurs when the force of gravity pulling the object downward is balanced by the force of air resistance pushing upward.Here's why terminal velocity matters:

• It means the object stops accelerating and falls at a constant speed.
• For many objects, terminal velocity is reached very quickly once falling begins.
• An object's shape, size, and mass affect its terminal velocity.

In the baseball's case, terminal velocity can be calculated using the formula:\[ v = \sqrt{\frac{2mg}{k}} \]- \( m \) is the mass of the baseball, which we've converted from ounces to pounds.- \( g \) still represents gravitational acceleration.- \( k \) is the drag coefficient.With the baseball reaching terminal velocity, it travels at a steady speed of approximately \( 106.87 \text{ft/s} \) when considering air resistance, compared to the faster speed of approximately \( 144.6 \text{ft/s} \) when ignoring air resistance. Thus, terminal velocity embodies the maximum speed the baseball can achieve under the balance of forces.