What is the reaction time for a batter facing a 45.0 m/s fastball?

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SUMMARY

The discussion centers on calculating the reaction time for a batter facing a fastball thrown at 45.0 m/s from a distance of 17.1 meters. The batter has a total reaction and swing time of 0.340 seconds. By applying the formula for uniform motion, the horizontal distance is divided by the velocity, resulting in a travel time of 0.04 seconds for the ball. This indicates that the batter has 0.04 seconds to react after the ball has left the pitcher's hand.

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Homework Statement


A baseball pitcher throws a fastball horizontally at a speed of 45.0 m/s toward home plate,
17.1 m away. Ignore air resistance. (a) If the batter’s combined reaction and swing times total
0.340 s, how long can the batter watch the ball after it has left the pitcher’s hand before
deciding to swing?


Homework Equations


I'm thinking x=x(not) + v(not)t + (1/2)at^2


The Attempt at a Solution


This equation seems perfect but I can't figure out how to solve for t so that I can find the time it takes the ball to travel the 17.1meters. Then I would take that time and substract the .340s from it and get the answer.

Thanks!
 
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Is the horizontal motion accelerated?
 
voko said:
Is the horizontal motion accelerated?

wow...thank you.

it is not. there for it is simply distance divided by velocity. giving the batter .04 seconds to react.
 

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