6) Bob wants to measure the height of a cliff. Unfortunately, he only has with him a stopwatch and a baseball. Luckily, he has a plan. While Jane stays at the bottom of the cliff, Bob climbs up the cliff face. When he reaches the top, he throws the baseball as hard as he can, with a velocity of 33.7m/s at an angle of 22.5°. Some time later, Jane shouts that the ball landed 127 m from the bottom of the cliff.

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### Problem: Measuring the Height of a Cliff

6) Bob wants to measure the height of a cliff. Unfortunately, he only has with him a stopwatch and a baseball. Luckily, he has a plan. While Jane stays at the bottom of the cliff, Bob climbs up the cliff face. When he reaches the top, he throws the baseball as hard as he can, with a velocity of 33.7 m/s at an angle of 22.5°. Some time later, Jane shouts that the ball landed 127 m from the bottom of the cliff.

    a) If Bob threw the ball from 1.50 m above the clifftop, how high is the cliff itself?

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This problem involves using physics, specifically kinematics, to determine the height of a cliff based on given parameters of the thrown ball. The setup involves determining the vertical displacement caused by the baseball being thrown from a known height and angle. 

#### Explanation:

- **Initial Velocity (v₀)**: 33.7 m/s
- **Angle of Projection (θ)**: 22.5°
- **Horizontal Distance (d)**: 127 m
- **Initial Height of Throw (h₀)**: 1.50 m above the clifftop

##### Steps to Solve:
1. **Resolve the initial velocity into horizontal (v₀x) and vertical (v₀y) components:**
   - \( v₀x = v₀ \cdot \cos(\theta) \)
   - \( v₀y = v₀ \cdot \sin(\theta) \)

2. **Calculate the time of flight (t) using the horizontal motion:**
   - \( d = v₀x \cdot t \)
   - \( t = \frac{d}{v₀x} \)

3. **Calculate the vertical displacement (y) using the vertical motion:**
   - Equation of motion: \( y = v₀y \cdot t - \frac{1}{2} g t^2 \)
   - g (acceleration due to gravity) ≈ 9.8 m/s²

4. **Determine the total height of the cliff:**
   - Total height (H) = displacement \(\Delta y\) + initial height (h₀)

By substituting the above-calculated values into their respective equations, the height of
Transcribed Image Text:--- ### Problem: Measuring the Height of a Cliff 6) Bob wants to measure the height of a cliff. Unfortunately, he only has with him a stopwatch and a baseball. Luckily, he has a plan. While Jane stays at the bottom of the cliff, Bob climbs up the cliff face. When he reaches the top, he throws the baseball as hard as he can, with a velocity of 33.7 m/s at an angle of 22.5°. Some time later, Jane shouts that the ball landed 127 m from the bottom of the cliff. a) If Bob threw the ball from 1.50 m above the clifftop, how high is the cliff itself? --- This problem involves using physics, specifically kinematics, to determine the height of a cliff based on given parameters of the thrown ball. The setup involves determining the vertical displacement caused by the baseball being thrown from a known height and angle. #### Explanation: - **Initial Velocity (v₀)**: 33.7 m/s - **Angle of Projection (θ)**: 22.5° - **Horizontal Distance (d)**: 127 m - **Initial Height of Throw (h₀)**: 1.50 m above the clifftop ##### Steps to Solve: 1. **Resolve the initial velocity into horizontal (v₀x) and vertical (v₀y) components:** - \( v₀x = v₀ \cdot \cos(\theta) \) - \( v₀y = v₀ \cdot \sin(\theta) \) 2. **Calculate the time of flight (t) using the horizontal motion:** - \( d = v₀x \cdot t \) - \( t = \frac{d}{v₀x} \) 3. **Calculate the vertical displacement (y) using the vertical motion:** - Equation of motion: \( y = v₀y \cdot t - \frac{1}{2} g t^2 \) - g (acceleration due to gravity) ≈ 9.8 m/s² 4. **Determine the total height of the cliff:** - Total height (H) = displacement \(\Delta y\) + initial height (h₀) By substituting the above-calculated values into their respective equations, the height of
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