5. A cannon fires a projectile perfectly horizontal with initial speed of vi = 40m/s. The cannon Quinontol dictonce R from the launch position will the proiectile hit the ground? For sim

Physics for Scientists and Engineers: Foundations and Connections
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Author:Katz, Debora M.
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**Problem 5: Horizontal Projectile Motion**

*A cannon fires a projectile perfectly horizontal with an initial speed of \( v_i = 40 \, \text{m/s} \). The cannon is at an initial height of \( H = 5 \, \text{m} \). At what horizontal distance \( R \) from the launch position will the projectile hit the ground? For simplicity use \( g = 10 \, \text{m/s}^2 \).*

---

**Solution**

To solve this problem, use the following steps:

1. **Calculate the time of flight:**

   Since the projectile is launched horizontally, the time \( t \) it takes to hit the ground depends only on the vertical motion.

   \[
   H = \frac{1}{2}gt^2
   \]

   \[
   5 = \frac{1}{2}(10)t^2
   \]

   \[
   5 = 5t^2
   \]

   \[
   t^2 = 1 \quad \Rightarrow \quad t = 1 \, \text{second}
   \]

2. **Calculate the horizontal distance \( R \):**

   The horizontal distance is calculated using the horizontal speed and the time of flight:

   \[
   R = v_i \times t
   \]

   \[
   R = 40 \, \text{m/s} \times 1 \, \text{s}
   \]

   \[
   R = 40 \, \text{meters}
   \]

Therefore, the projectile will hit the ground at a horizontal distance of 40 meters from the launch position.
Transcribed Image Text:**Problem 5: Horizontal Projectile Motion** *A cannon fires a projectile perfectly horizontal with an initial speed of \( v_i = 40 \, \text{m/s} \). The cannon is at an initial height of \( H = 5 \, \text{m} \). At what horizontal distance \( R \) from the launch position will the projectile hit the ground? For simplicity use \( g = 10 \, \text{m/s}^2 \).* --- **Solution** To solve this problem, use the following steps: 1. **Calculate the time of flight:** Since the projectile is launched horizontally, the time \( t \) it takes to hit the ground depends only on the vertical motion. \[ H = \frac{1}{2}gt^2 \] \[ 5 = \frac{1}{2}(10)t^2 \] \[ 5 = 5t^2 \] \[ t^2 = 1 \quad \Rightarrow \quad t = 1 \, \text{second} \] 2. **Calculate the horizontal distance \( R \):** The horizontal distance is calculated using the horizontal speed and the time of flight: \[ R = v_i \times t \] \[ R = 40 \, \text{m/s} \times 1 \, \text{s} \] \[ R = 40 \, \text{meters} \] Therefore, the projectile will hit the ground at a horizontal distance of 40 meters from the launch position.
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