One of the fastest recorded pitches in major league baseball, thrown by Nolan Ryan in 1974, was clocked at 100.8 mi/hr. If a pitch were thrown horizontally with this velocity, how far would the ball fall vertically by the time it reached home plate, 60.0 ft away?

The ball would fall vertically by approximately 2.10 feet by the time it reaches home plate 60.0 feet away when thrown horizontally at 100.8 mi/hr.

How to find how far would the ball fall vertically by the time it reached home plate 60.0 ft away

To find how far the ball would fall vertically by the time it reaches home plate, we can use the equations of motion. We'll first convert the velocity from miles per hour (mi/hr) to feet per second (ft/s) because the distance is given in feet.

First, convert the velocity:

1 mile = 5280 feet

1 hour = 3600 seconds

So, \(v\) in ft/s is:

[tex]\[ v = \frac{100.8 \, \text{mi/hr} \times 5280 \, \text{ft/mi}}{3600 \, \text{s/hr}} \approx 147.2 \, \text{ft/s} \][/tex]

Now, we can use the following kinematic equation to find the vertical distance (\(y\)) the ball falls:

[tex]\[ y = \frac{1}{2}gt^2 \][/tex]

We need to find \(t\), which we can calculate using the horizontal distance (\(d\)) and the horizontal velocity (\(v\)):

[tex]\[ t = \frac{d}{v} = \frac{60.0 \, \text{ft}}{147.2 \, \text{ft/s}} \approx 0.407 \, \text{s} \][/tex]

Now, substitute \(t\) into the vertical motion equation:

[tex]\[ y = \frac{1}{2} \cdot 32.2 \, \text{ft/s}^2 \cdot (0.407 \, \text{s})^2 \][/tex]

Calculate \(y\):

[tex]\[ y \approx 2.10 \, \text{ft} \][/tex]

So, the ball would fall vertically by approximately 2.10 feet by the time it reaches home plate 60.0 feet away when thrown horizontally at 100.8 mi/hr.

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