College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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**Physics Problem: Projectile Motion**

**Question:**
Rocky the Flying Squirrel is carrying a nut of mass 0.5 kg while flying horizontally at a height of 15 m above the ground at a speed of 12 m/s. Bullwinkle is eagerly awaiting the delivery of the nut on the ground. Rocky releases the nut as he is directly above Bullwinkle. How far from Bullwinkle will the nut land if Bullwinkle does not move?

**Options:**
- ○ 8.49 m
- ○ 5.20 m
- ○ 4.24 m
- ○ 20.8 m

**Explanation:**
To solve this question, you need to apply the principles of projectile motion. Here's a step-by-step guide:

1. **Determine the time taken for the nut to fall to the ground:**
   - The initial vertical velocity is 0 m/s (since it's only moving horizontally).
   - Use the equation for the vertical displacement due to gravity:
     \[
     h = \frac{1}{2} g t^2
     \]
     where \( h = 15 \, \text{m} \) is the height and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

     Solving for \( t \):
     \[
     15 = \frac{1}{2} \times 9.8 \times t^2
     \]
     \[
     t^2 = \frac{30}{9.8} \approx 3.06
     \]
     \[
     t \approx 1.75 \, \text{s}
     \]

2. **Calculate the horizontal distance traveled by the nut:**
   - The horizontal velocity remains constant at 12 m/s.
   - The time of flight calculated above is \( t \approx 1.75 \) s.

     Use the formula for horizontal distance:
     \[
     d = v \times t
     \]
     where \( v = 12 \, \text{m/s} \).

     Substituting the values:
     \[
     d = 12 \times 1.75 \approx 21 \, \text{m}
     \]

Since the closest answer to this calculation is 20.8 m, the correct option is:

- ○
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Transcribed Image Text:**Physics Problem: Projectile Motion** **Question:** Rocky the Flying Squirrel is carrying a nut of mass 0.5 kg while flying horizontally at a height of 15 m above the ground at a speed of 12 m/s. Bullwinkle is eagerly awaiting the delivery of the nut on the ground. Rocky releases the nut as he is directly above Bullwinkle. How far from Bullwinkle will the nut land if Bullwinkle does not move? **Options:** - ○ 8.49 m - ○ 5.20 m - ○ 4.24 m - ○ 20.8 m **Explanation:** To solve this question, you need to apply the principles of projectile motion. Here's a step-by-step guide: 1. **Determine the time taken for the nut to fall to the ground:** - The initial vertical velocity is 0 m/s (since it's only moving horizontally). - Use the equation for the vertical displacement due to gravity: \[ h = \frac{1}{2} g t^2 \] where \( h = 15 \, \text{m} \) is the height and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. Solving for \( t \): \[ 15 = \frac{1}{2} \times 9.8 \times t^2 \] \[ t^2 = \frac{30}{9.8} \approx 3.06 \] \[ t \approx 1.75 \, \text{s} \] 2. **Calculate the horizontal distance traveled by the nut:** - The horizontal velocity remains constant at 12 m/s. - The time of flight calculated above is \( t \approx 1.75 \) s. Use the formula for horizontal distance: \[ d = v \times t \] where \( v = 12 \, \text{m/s} \). Substituting the values: \[ d = 12 \times 1.75 \approx 21 \, \text{m} \] Since the closest answer to this calculation is 20.8 m, the correct option is: - ○
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