2) Your nephew drops a stuffed bunny from the second floor balcony. You want to get the bunny back to him so you toss it up to the second floor which is 3.8 meters above you. What is the minimum velocity you need to toss the bunny to reach the second floor (ignore air resistance)? (15 pts)

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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Problem Scenario:**

Your nephew drops a stuffed bunny from the second-floor balcony. You want to get the bunny back to him, so you toss it up to the second floor, which is 3.8 meters above you. 

**Question:**

What is the minimum velocity you need to toss the bunny to reach the second floor (ignore air resistance)? 

**Points:**

15 points

---

**Solution Approach:**

To solve this problem, you can use the principles of physics related to projectile motion, specifically focusing on the kinematic equations. Ignoring air resistance simplifies the calculations:

1. **Identify the knowns:**
   - Height to reach: 3.8 meters
   - Initial velocity (\(v_0\)): ? (This is what we need to find)
   - Final velocity (\(v\)): 0 m/s (at the peak of the trajectory for minimum velocity)
   - Acceleration (\(a\)): \(-9.8 \, \text{m/s}^2\) (due to gravity, acting downwards)

2. **Use the kinematic equation:**

\[
v^2 = v_0^2 + 2a \cdot s
\]

where:
- \(v\) is the final velocity (0 m/s at the peak),
- \(v_0\) is the initial velocity,
- \(a\) is the acceleration due to gravity \(-9.8 \, \text{m/s}^2\),
- \(s\) is the distance (3.8 meters).

3. **Rearrange to find the initial velocity:**

\[
0 = v_0^2 + 2(-9.8) \cdot 3.8
\]

\[
v_0^2 = 2 \cdot 9.8 \cdot 3.8
\]

\[
v_0 = \sqrt{2 \cdot 9.8 \cdot 3.8}
\]

4. **Calculate \(v_0\):**

- Plug in the values and compute \(v_0\) to find the minimum velocity required to reach the second floor.
Transcribed Image Text:**Problem Scenario:** Your nephew drops a stuffed bunny from the second-floor balcony. You want to get the bunny back to him, so you toss it up to the second floor, which is 3.8 meters above you. **Question:** What is the minimum velocity you need to toss the bunny to reach the second floor (ignore air resistance)? **Points:** 15 points --- **Solution Approach:** To solve this problem, you can use the principles of physics related to projectile motion, specifically focusing on the kinematic equations. Ignoring air resistance simplifies the calculations: 1. **Identify the knowns:** - Height to reach: 3.8 meters - Initial velocity (\(v_0\)): ? (This is what we need to find) - Final velocity (\(v\)): 0 m/s (at the peak of the trajectory for minimum velocity) - Acceleration (\(a\)): \(-9.8 \, \text{m/s}^2\) (due to gravity, acting downwards) 2. **Use the kinematic equation:** \[ v^2 = v_0^2 + 2a \cdot s \] where: - \(v\) is the final velocity (0 m/s at the peak), - \(v_0\) is the initial velocity, - \(a\) is the acceleration due to gravity \(-9.8 \, \text{m/s}^2\), - \(s\) is the distance (3.8 meters). 3. **Rearrange to find the initial velocity:** \[ 0 = v_0^2 + 2(-9.8) \cdot 3.8 \] \[ v_0^2 = 2 \cdot 9.8 \cdot 3.8 \] \[ v_0 = \sqrt{2 \cdot 9.8 \cdot 3.8} \] 4. **Calculate \(v_0\):** - Plug in the values and compute \(v_0\) to find the minimum velocity required to reach the second floor.
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