A major league pitcher can throw a baseball in excess of 41.3 m/s.

If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches the catcher who is 16.1 m away from the point of release?

To determine how much the ball will drop by the time it reaches the catcher, we need to consider the horizontal distance traveled, and then use the equations of motion to calculate the vertical displacement.

First, let's find the time it takes for the ball to reach the catcher. We know the horizontal distance traveled is 16.1 m, and the initial horizontal velocity is 41.3 m/s. Since the ball is thrown horizontally, there will be no horizontal acceleration. Therefore, we can use the formula:

time = distance / velocity

time = 16.1 m / 41.3 m/s
time ≈ 0.3898 s

Next, we can determine the vertical displacement (drop) of the ball during this time. The ball will experience vertical acceleration due to gravity, which is approximately 9.8 m/s². We can use the formula:

displacement = (1/2) * acceleration * time²

displacement = (1/2) * 9.8 m/s² * (0.3898 s)²
displacement ≈ 0.7274 m

Therefore, the ball will drop approximately 0.7274 meters by the time it reaches the catcher.