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Chapter 2: Problem 8

In a game against the White Sox, baseball pitcher Nolan Ryan threw a pitch measured at \(45.1 \mathrm{m} / \mathrm{s} .\) If it was \(18.4 \mathrm{m}\) from Nolan's position on the pitcher's mound to home plate, how long did it take the ball to get to the batter waiting at home plate? Treat the ball's velocity as constant and ignore any gravitational effects.

Short Answer

Expert verified
Answer: Approximately 0.408 seconds.

Step by step solution

01

Write down the given information

We are given the following information: - Velocity of the baseball (\(v\)): \(45.1 \mathrm{m/s}\) - Distance between the pitcher's mound and home plate (\(d\)): \(18.4 \mathrm{m}\) Our goal is to find the time it takes (\(t\)) for the ball to travel from the pitcher's mound to home plate.
02

Use the formula for constant velocity

We will use the formula for constant velocity, which is: \(distance = velocity \times time\) In our case, the distance (\(d\)) is \(18.4 \mathrm{m}\) and the velocity (\(v\)) is \(45.1 \mathrm{m/s}\). So, we can rewrite the formula as: \(18.4 \mathrm{m} = 45.1 \mathrm{m/s} \times t\)
03

Solve for time

To solve for the time (\(t\)), we will divide both sides of the equation by the velocity: \(t = \frac{18.4 \mathrm{m}}{45.1 \mathrm{m/s}}\) By calculating the fraction, we get: \(t = 0.408 \mathrm{s}\)
04

State the final answer

The time it takes for the baseball to travel from the pitcher's mound to home plate is approximately \(0.408 \mathrm{s}\).

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Most popular questions from this chapter

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