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Chapter 2: Problem 14

A ball thrown by a pitcher on a women's softball team is timed at 65.0 mph. The distance from the pitching rubber to home plate is \(43.0 \mathrm{ft}\). In major league baseball the corresponding distance is \(60.5 \mathrm{ft}\). If the batter in the softball game and the batter in the baseball game are to have equal times to react to the pitch, with what speed must the baseball be thrown? Assume the ball travels with a constant velocity. [Hint: There is no need to convert units; set up a ratio.]

Short Answer

Expert verified
Answer: The baseball must be thrown at approximately 93 mph for the batters in both games to have an equal reaction time.

Step by step solution

01

Write down the given information for the softball pitch

We are given: - Softball pitch speed: \(65.0 \mathrm{mph}\) - Softball pitch distance: \(43.0 \mathrm{ft}\)
02

Write down the given information for the baseball pitch

We are given: - Baseball pitch distance: \(60.5 \mathrm{ft}\) - Baseball pitch speed: Unknown (\(x\))
03

Use the velocity formula to find the time of the softball pitch

Since we have the speed and distance for the softball pitch, we can use the formula for the velocity \(v_s = \frac{d_s}{t_s}\), where \(v_s\) is the softball pitch speed, \(d_s\) is the softball pitch distance, and \(t_s\) is the softball pitch time. Rearrange the formula to solve for \(t_s\): \(t_s = \frac{d_s}{v_s}\)
04

Use the velocity formula to find the time of the baseball pitch

Since we have the distance for the baseball pitch and we want the reaction times to be equal, we can use the formula for the velocity \(v_b = \frac{d_b}{t_b}\), where \(v_b\) is the baseball pitch speed, \(d_b\) is the baseball pitch distance and \(t_b\) is the baseball pitch time. Since we want the reaction times to be equal, we know that \(t_s = t_b\). Thus, we can rewrite the velocity formula for the baseball pitch as: \(v_b = \frac{d_b}{t_s}\)
05

Set up a ratio using the velocity formulas

Now, we can set up a ratio using the softball and baseball pitch velocity formulas. \(\frac{d_s}{v_s} = \frac{d_b}{v_b}\)
06

Substitute the given values and solve for the baseball pitch speed

Substitute the given values into the equation and solve for \(v_b\): \(\frac{43.0 \mathrm{ft}}{65.0 \mathrm{mph}} = \frac{60.5 \mathrm{ft}}{v_b}\) Cross-multiply to obtain: \(43.0 \mathrm{ft} \times v_b = 65.0 \mathrm{mph} \times 60.5 \mathrm{ft}\) Now, divide by \(43.0 \mathrm{ft}\): \(v_b = \frac{65.0 \mathrm{mph} \times 60.5 \mathrm{ft}}{43.0 \mathrm{ft}}\) \(v_b \approx 93.023 \mathrm{mph}\) The baseball must be thrown at approximately 93 mph for the batters in both games to have an equal reaction time.

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Most popular questions from this chapter

In the problems, please assume the free-fall acceleration $g=9.80 \mathrm{m} / \mathrm{s}^{2}$ unless a more precise value is given in the problem statement. Ignore air resistance. A stone is thrown vertically downward from the roof of a building. It passes a window \(16.0 \mathrm{m}\) below the roof with a speed of $25.0 \mathrm{m} / \mathrm{s} .\( It lands on the ground \)3.00 \mathrm{s}$ after it was thrown. What was (a) the initial velocity of the stone and (b) how tall is the building?
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