anonymous
  • anonymous
A fast base-runner can get from first to second base in 3.46 s. If he leaves first base as the pitcher throws a 90 mi/h fastball the 68-ft distance to the catcher, and if the catcher takes 0.45 s to catch and rethrow the ball, how fast does the catcher have to throw the ball to second base to make an out? Home plate to second base is the diagonal of a square 90 ft on a side.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
a fig would help.
anonymous
  • anonymous
this is the only information my professor gave me
Shane_B
  • Shane_B
You don't really need a figure for this one. Here's how I'd do it: 1) Convert 90 mph to ft/s:\[\frac{90mi}{1hr}*\frac{5280ft}{3600s}=132ft/s\] 2) Determine how long it takes for the pitch to travel 68ft and get to the catcher: \[\frac{68ft}{132ft/s}=0.52s\] 3) Subtract the pitch time and catcher delay from the time it takes the runner to get to 2nd base:\[3.46s-0.45s-0.52s=2.49s\]So the catcher needs to get the ball to second base in under 2.49s since a tie goes to the runner. 4) Determine how far the catcher needs to throw the ball. Using the Pythagorean Theorem, the distance to from the catcher to 2nd base will be \[d=\sqrt{90^2+90^2}=127.3ft\] Finally, calculate how fast he needs to throw the ball: \[\frac{127.3ft}{2.49s}=51.12ft/s\]That's about 34.86 mph...so he needs to throw the ball slightly faster than that so that he beats the runner.

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anonymous
  • anonymous
Thanks so much!!
Shane_B
  • Shane_B
np :)

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