Final answer:
The ball would fall vertically by approximately 16.07 feet by the time it reached home plate.
Step-by-step explanation:
To determine the distance the baseball would fall vertically by the time it reached home plate, we can use the equations of motion. Since the pitch is thrown horizontally, the initial vertical velocity is 0. The time taken for the ball to reach home plate is the same as the time taken for the ball to travel horizontally, which can be calculated using the velocity and distance traveled horizontally.
Using the equation d = vxt, where d is the distance, vx is the horizontal velocity, and t is the time, we can rearrange the equation to solve for t. By dividing the distance traveled horizontally by the horizontal velocity, we get the time taken.
Once we have the time, we can use the equation d = 1/2gt2, where d is the distance fallen vertically, g is the acceleration due to gravity (approximately 9.8 m/s2), and t is the time, to calculate the distance fallen vertically. Converting this distance to feet, we get the final answer.
In this case, the horizontal velocity is 101.0 mi/h and the distance traveled horizontally is 60.5 ft. So, the time taken is 60.5 ft / 101.0 mi/h = 1 s (since 1 mi/h = 1.46667 ft/s). Using this value of time in the equation d = 1/2gt2, we get d = 1/2 * 9.8 m/s2 * (1 s)^2 = 4.9 m = 16.07 ft (approximately). Therefore, the ball would fall vertically by approximately 16.07 feet by the time it reached home plate.