A softball having a mass of 0.25kg is pitched at 86.4km/h. By the time it reaches the plate, it may have slowed by 11%. Neglecting gravity, estimate the average force of air resistance during a pitch, if the distance between the plate and the pitcher is about 15m.

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### Problem Statement

A softball having a mass of 0.25kg is pitched at 86.4km/h. By the time it reaches the plate, it may have slowed by 11%. Neglecting gravity, estimate the average force of air resistance during a pitch, if the distance between the plate and the pitcher is about 15m.

### Explanation

In this problem, we need to calculate the average force of air resistance acting on a softball during a pitch. Here are the given data and steps to find the solution:

- Mass of the softball, \( m = 0.25 \) kg
- Initial speed of the softball, \( v_i = 86.4 \) km/h (converted to m/s: \( 86.4 \div 3.6 = 24 \) m/s)
- The speed reduction is 11%, so the final speed is \( v_f = 24 \times (1 - 0.11) = 21.36 \) m/s
- Distance covered by the softball, \( d = 15 \) m

To find the average force of air resistance:

1. **Calculate deceleration (a):**
   Using the kinematic equation \( v_f^2 = v_i^2 + 2ad \):
   \[
   (21.36)^2 = (24)^2 + 2a(15)
   \]
   Solving for \( a \):
   \[
   456.2496 = 576 + 30a
   \]
   \[
   30a = 456.2496 - 576
   \]
   \[
   a = -3.9734 \: \text{m/s}^2
   \]

2. **Calculate the average force of air resistance (F):**
   Using Newton’s second law \( F = ma \):
   \[
   F = 0.25 \times (-3.9734)
   \]
   \[
   F \approx -0.9934 \: \text{N}
   \]

The negative sign indicates that the force acts in the direction opposite to the motion of the ball.

### Summary

Thus, the average force of air resistance acting on the softball during its pitch is approximately \( 0.9934 \) N.
Transcribed Image Text:### Problem Statement A softball having a mass of 0.25kg is pitched at 86.4km/h. By the time it reaches the plate, it may have slowed by 11%. Neglecting gravity, estimate the average force of air resistance during a pitch, if the distance between the plate and the pitcher is about 15m. ### Explanation In this problem, we need to calculate the average force of air resistance acting on a softball during a pitch. Here are the given data and steps to find the solution: - Mass of the softball, \( m = 0.25 \) kg - Initial speed of the softball, \( v_i = 86.4 \) km/h (converted to m/s: \( 86.4 \div 3.6 = 24 \) m/s) - The speed reduction is 11%, so the final speed is \( v_f = 24 \times (1 - 0.11) = 21.36 \) m/s - Distance covered by the softball, \( d = 15 \) m To find the average force of air resistance: 1. **Calculate deceleration (a):** Using the kinematic equation \( v_f^2 = v_i^2 + 2ad \): \[ (21.36)^2 = (24)^2 + 2a(15) \] Solving for \( a \): \[ 456.2496 = 576 + 30a \] \[ 30a = 456.2496 - 576 \] \[ a = -3.9734 \: \text{m/s}^2 \] 2. **Calculate the average force of air resistance (F):** Using Newton’s second law \( F = ma \): \[ F = 0.25 \times (-3.9734) \] \[ F \approx -0.9934 \: \text{N} \] The negative sign indicates that the force acts in the direction opposite to the motion of the ball. ### Summary Thus, the average force of air resistance acting on the softball during its pitch is approximately \( 0.9934 \) N.
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