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One of the fastest recorded pitches in major league bascball, thrown by Tim Lincecum in 2009 , was clocked at 101.0 $\mathrm{mi} / \mathrm{h}$ (Fig. P3.8). If a pitch were thrown horizontally with this velocity, how far would the ball fall vertically by the time it reached home plate, 60.5 $\mathrm{ft}$ away?

$y = - 0.82 \mathrm { m } = - 2.7 \mathrm { ft }$

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Cornell University

Numerade Educator

Hope College

McMaster University

we know that the Auto X is gonna be equaling the X times t There isn't city eggs acceleration in the ex direction. So that means that tea is gonna be equal in Delta X, divided by the visa back the velocity of the extraction. And so that this time it's gonna be equaling. This would be 60 feet, 60.0 feet, multiplied by point 3048 meters per foot. And then this would be divided by 100 miles per hour, both supplied by 1609 meters for every mile multiplied by one hour for every 3600 seconds. And we find that time is equally point for one seconds. And we're going to then say that Delta Y equals V. Why initial t plus 1/2 g t squared. And in this case, we can say that then Delta, we know that the initial velocity is gonna be zero. And so we can say that the change and why it's going to be equally 1/2 times 9.8 meters per second squared both supplied my point for one second's quantity squared. This is giving us negative point, brother it'll be positive 0.82 meters, but downwards cause we chose downwards to be positive in this case. So this would be how far the ball falls vertically. So 0.82 meters down birds. That is the end of the solution. Thank you for watching.