Trying to escape his pursuers, a secret agent skis off a slope inclined at 30° below the horizontal at 60 km/h. To survive and land on the snow 140 m below, he/she must clear a gorge 60 m wide. Does he/she make it? Ignore air resistance.

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Author:Paul W. Zitzewitz
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Chapter6: Motion In Two Dimensions
Section: Chapter Questions
Problem 90A
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Does the skier clear the 60 m gorge
**Physics Problem: Skiing Stunt - Projectile Motion**

**Problem Statement:**

Trying to escape their pursuers, a secret agent skis off a slope inclined at 30° below the horizontal at 60 km/h. To survive and land on the snow 140 m below, they must clear a gorge 60 m wide. Does the secret agent make it? Ignore air resistance.

**Explanation of Diagram:**

1. **Inclined Slope:**
   - The secret agent is shown skiing off the edge of a slope that is inclined at a 30° angle below the horizontal.

2. **Initial Velocity (v₀):**
   - The agent’s initial velocity as they leave the slope is 60 km/h.

3. **Gorge Dimensions:**
   - The vertical distance to clear is 140 meters.
   - The horizontal distance to clear is 60 meters.

4. **Important Points:**
   - Label showing "Edge" where the agent starts skiing off.
   - Arrow indicating the direction of the initial velocity (v₀).
   - Notes indicating that the diagram is not to scale.

**Objective:**

Determine if the secret agent can successfully land on the other side of the gorge by calculating the horizontal range of their projectile motion, considering the given initial velocity and angles.

**Required Calculations:**

To solve this problem, we need to:
1. Resolve the initial velocity into horizontal (v₀x) and vertical (v₀y) components.
2. Use kinematic equations to find the time of flight (t) based on vertical motion.
3. Calculate the horizontal range (R) using the time of flight.

**Formulas:**

1. Initial velocity components:
   - \( v₀x = v₀ \cdot \cos(30°) \)
   - \( v₀y = v₀ \cdot \sin(30°) \)

2. Time of flight (t) from vertical motion:
   - Use the formula for vertical displacement under gravity:
     \[ y = v₀y \cdot t + \frac{1}{2} g t² \]
     Here, \( y = -140 m \), \( g = 9.8  m/s² \) (acceleration due to gravity)

3. Horizontal range (R):
   - \( R = v₀x \cdot t \)

By plugging in
Transcribed Image Text:**Physics Problem: Skiing Stunt - Projectile Motion** **Problem Statement:** Trying to escape their pursuers, a secret agent skis off a slope inclined at 30° below the horizontal at 60 km/h. To survive and land on the snow 140 m below, they must clear a gorge 60 m wide. Does the secret agent make it? Ignore air resistance. **Explanation of Diagram:** 1. **Inclined Slope:** - The secret agent is shown skiing off the edge of a slope that is inclined at a 30° angle below the horizontal. 2. **Initial Velocity (v₀):** - The agent’s initial velocity as they leave the slope is 60 km/h. 3. **Gorge Dimensions:** - The vertical distance to clear is 140 meters. - The horizontal distance to clear is 60 meters. 4. **Important Points:** - Label showing "Edge" where the agent starts skiing off. - Arrow indicating the direction of the initial velocity (v₀). - Notes indicating that the diagram is not to scale. **Objective:** Determine if the secret agent can successfully land on the other side of the gorge by calculating the horizontal range of their projectile motion, considering the given initial velocity and angles. **Required Calculations:** To solve this problem, we need to: 1. Resolve the initial velocity into horizontal (v₀x) and vertical (v₀y) components. 2. Use kinematic equations to find the time of flight (t) based on vertical motion. 3. Calculate the horizontal range (R) using the time of flight. **Formulas:** 1. Initial velocity components: - \( v₀x = v₀ \cdot \cos(30°) \) - \( v₀y = v₀ \cdot \sin(30°) \) 2. Time of flight (t) from vertical motion: - Use the formula for vertical displacement under gravity: \[ y = v₀y \cdot t + \frac{1}{2} g t² \] Here, \( y = -140 m \), \( g = 9.8 m/s² \) (acceleration due to gravity) 3. Horizontal range (R): - \( R = v₀x \cdot t \) By plugging in
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