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Chapter 9: Problem 30

A typical Major League fastball is thrown at approximately \(88 \mathrm{mph}\) and with a spin rate of \(110 \mathrm{rpm} .\) If the distance between the pitcher's point of release and the catcher's glove is exactly \(60.5 \mathrm{ft},\) how many full turns does the ball make between release and catch? Neglect any effect of gravity or air resistance on the ball's flight.

Short Answer

Expert verified
Answer: 0 full turns

Step by step solution

01

Convert mph to ft/s

Convert the velocity of the fastball from 88 mph to feet per second (ft/s). 1 mile = 5280 feet and 1 hour = 3600 seconds. So, we have the following conversion formula: \(88\frac{\text{miles}}{\text{hour}} \times \frac{5280\text{ feet}}{1\text{ mile}} \times \frac{1\text{ hour}}{3600\text{ seconds}}.\)
02

Calculate the fastball's velocity in ft/s

Now, calculate the fastball's velocity in ft/s. \(88 \frac{\text{miles}}{\text{hour}} \times \frac{5280\text{ feet}}{1\text{ mile}} \times \frac{1\text{ hour}}{3600\text{ seconds}}\approx 128.96\text{ ft/s}.\)
03

Convert rpm to turns/s

Convert the spin rate of the fastball from 110 rpm to turns per second (turns/s). 1 minute = 60 seconds. So, we have the following conversion formula: \(110\frac{\text{turns}}{\text{minute}} \times \frac{1\text{ minute}}{60\text{ seconds}}.\)
04

Calculate the spin rate in turns/s

Now, calculate the spin rate of the fastball in turns/s. \(110\frac{\text{turns}}{\text{minute}} \times \frac{1\text{ minute}}{60\text{ seconds}}\approx 1.83\text{ turns/s}\)
05

Find the travel time

Calculate the time it takes for the fastball to travel the given distance of 60.5 ft. Speed = distance / time Re-arrange for time: time = distance / speed \(\text{time} = \frac{60.5\text{ ft}}{128.96\text{ ft/s}}\approx 0.47\text{ seconds}\)
06

Calculate the total turns

Multiply the time it takes for the ball to travel the distance by the spin rate in turns/s to find out how many total turns the ball makes between release and catch. Total turns = turns/s × time \(\text{Total turns} = 1.83\text{ turns/s} \times 0.47\text{ seconds} \approx 0.86\text{ turns}\). Since only a whole number of full turns can be made, the number of full turns that the ball makes between release and catch is 0.

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