3. To predict theoretically the new range, x, of a ball shot off with an initial velocity at angle of 55.0 degrees above the horizontal, first predict the time of flight using the formula for the vertical equation and then compute the horizontal range x y = (v, sin e )t - 1/2gt? x=(v, cos 0 ) t Show your work.

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Can you please help me with number 3? Using the equations given. Thank you!
**Transcription of the Educational Document**

**Title: Physics Experiment on Projectile Motion**

**Vertical Distance of Fall:** 1.13 m (5 ft.)

**Table: Recorded Distances**
| Trial Number | 1    | 2    | 3    | 4    | 5    | 6    | 7    | 8    | 9    | 10   | Average |
|--------------|------|------|------|------|------|------|------|------|------|------|---------|
| Distance (m) | 1.74 | 1.74 | 1.74 | 1.74 | 1.73 | 1.73 | 1.74 | 1.73 | 1.74 | 1.74 | 1.738  |

**1. Calculating the Average Value:**
- The average value of the data was calculated by summing all trial distances and dividing by the number of trials.
  
**2. Initial Velocity Calculation:**
- Using the equation of projectile motion, compute the initial velocity of the projection. Show your work:
  - \( v = \sqrt{2gh} \)
  - Calculations include:
    - \( v = 3.619 \, \text{m/s} \)
    - Another value calculated is \( u = 3.03 \, \text{m/s} \)

**3. Adjusting Launch Angle to 55.0 Degrees:**
- The student adjusted the angle of the projectile launcher to 55.0 degrees using the angle indicator on the side of the barrel.
- Using the initial velocity found in procedure 1 and the vertical distance measured, calculate the new time of flight and new horizontal distance.
- Assume the ball is shot off at the new angle.

**4. Predicting Theoretical Range, x:**
- To predict the new range \( x \) of a ball shot off with an initial velocity at an angle of 55.0 degrees:
  - First, predict the time of flight using the formula for vertical motion:
    - \( y = (v \sin \theta)t - \frac{1}{2}gt^2 \)
  - Then compute the horizontal range: 
    - \( x = (v \cos \theta)t \)

**Sample Work:**
- Calculating time of flight \( T \): 
  - \(
Transcribed Image Text:**Transcription of the Educational Document** **Title: Physics Experiment on Projectile Motion** **Vertical Distance of Fall:** 1.13 m (5 ft.) **Table: Recorded Distances** | Trial Number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Average | |--------------|------|------|------|------|------|------|------|------|------|------|---------| | Distance (m) | 1.74 | 1.74 | 1.74 | 1.74 | 1.73 | 1.73 | 1.74 | 1.73 | 1.74 | 1.74 | 1.738 | **1. Calculating the Average Value:** - The average value of the data was calculated by summing all trial distances and dividing by the number of trials. **2. Initial Velocity Calculation:** - Using the equation of projectile motion, compute the initial velocity of the projection. Show your work: - \( v = \sqrt{2gh} \) - Calculations include: - \( v = 3.619 \, \text{m/s} \) - Another value calculated is \( u = 3.03 \, \text{m/s} \) **3. Adjusting Launch Angle to 55.0 Degrees:** - The student adjusted the angle of the projectile launcher to 55.0 degrees using the angle indicator on the side of the barrel. - Using the initial velocity found in procedure 1 and the vertical distance measured, calculate the new time of flight and new horizontal distance. - Assume the ball is shot off at the new angle. **4. Predicting Theoretical Range, x:** - To predict the new range \( x \) of a ball shot off with an initial velocity at an angle of 55.0 degrees: - First, predict the time of flight using the formula for vertical motion: - \( y = (v \sin \theta)t - \frac{1}{2}gt^2 \) - Then compute the horizontal range: - \( x = (v \cos \theta)t \) **Sample Work:** - Calculating time of flight \( T \): - \(
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