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A major-league pitcher can throw a baseball in excess of 41.0 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is 17.0 m away from the point of release?

$h=0.842 \mathrm{m}$

Physics 101 Mechanics

Chapter 3

Kinematics in Two Dimensions

Motion in 2d or 3d

Cornell University

University of Michigan - Ann Arbor

Simon Fraser University

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so the question states that a pitcher can throw a ball at 41 meters per second and we're trying to find how far the ball falls after it moves 17 meters. Um so the first thing we really should do is find the time it takes to move 17 meters. To do this. We know that the velocity times the time is equal to distance and because we know the horrible horizontal velocity is 41 meters per second and we also know that distance we can solve for t. So 41 meters per second times time is equal to 17 meters. Divide by 41 on both sides and we get that the time is equal to 17 who? 41 seconds, which is just the same thing as point for one for six seconds. Now that we know the time that it takes to move 17 meters, we can just plug it into our kingdom attics formula which states that the change in displacement in the Y direction Let's call it Delta X, is equal to the initial velocity in the Y direction times, time plus 1/2 a T squared the initial velocity in this case in the Y direction is just zero, so you can give it to this term. And we know the acceleration in the Y direction is always gonna be negative 9.8 meters per second squared. We also know the time that we calculated down here. So when we plug everything in, we find that Delta X is equal to negative 0.842 meters and it's negative because it dropped in this value, so we can really just take the positive value. So the answer is 0.842 meters.

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