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Chapter 2: Problem 44

A base runner can get from first to second base in 3.4 s. If he leaves first as the pitcher throws a 90 milh fastball the 61 -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 \(90 \mathrm{ft}\) on a side.

Short Answer

Expert verified
The catcher has to throw the ball to second base at a speed of 50.91 mph to make an out.

Step by step solution

01

Calculate the distance from home plate to second base

Since home plate to second base is a diagonal of a square of side 90 ft, use the Pythagorean theorem to get the distance. In a square, the diagonal \(d\) can be calculated by \(d = \sqrt{2} * a\) where \(a\) is the side of the square. So, the diagonal or the distance from home plate to second base = \( \sqrt{2} * 90 ft = 127.28 ft \)
02

Calculate the total time for the base runner to reach the second base

The base runner to reach second base will take 3.4s. That's the total time the pitcher has, starting from when he throws the ball, including the catcher catching and re-throwing the ball. Hence, total time = 3.4 s
03

Calculate the time for ball to travel from pitcher to catcher and the catcher to throw the ball

Time for ball to travel from pitcher to catcher = Distance/Speed = 61ft / 90 mph = 0.45s. The catcher needs 0.45 seconds to catch and rethrow the ball. Therefore, total time spent before the catcher throws the ball = Time ball travels + catcher's time = 0.45s + 0.45s = 0.9s.
04

Calculate the time for ball to travel from catcher to second base

To calculate the time ball has to travel from catcher to second base, subtract the total time spent before the catcher throws the ball (calculated in Step 3) from the base runner's total time. Hence, Time for the ball to travel from catcher to second base = Total Time - Time before catcher throws = 3.4s - 0.9s = 2.5s.
05

Calculate the speed of the ball from catcher to second base

Finally, to determine the speed at which the catcher has to throw the ball to second base to make an out, the speed formula should be used again. Thus, Speed = Distance/Time. Substituting the given values, we find that the Speed of the ball from catcher to second base = 127.28 ft / 2.5 s = 50.91 mph. This means that the catcher has to throw the ball at a speed of 50.91 mph for it to reach the second base on time for an out.

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Most popular questions from this chapter

A jetliner leaves San Francisco for New York, \(4600 \mathrm{km}\) away. With a strong tailwind, its speed is \(1100 \mathrm{km} / \mathrm{h}\). At the same time, a second jet leaves New York for San Francisco. Flying into the wind, it makes only \(700 \mathrm{km} / \mathrm{h}\). When and where do the two planes pass?

You allow 40 min to drive 25 mi to the airport, but you're caught in heavy traffic and average only \(20 \mathrm{mi} / \mathrm{h}\) for the first 15 min. What must your average speed be on the rest of the trip if you're to make your flight?

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