An object reaches a maximum vertical height of 23.0 m when thrown vertically upward on the earth. How high would it travel on the moon where the acceleration due to gravity is about one sixth that on the earth ? Assume that initial velocity is the same.

Options

A. 138 m
B. 100 m
C. 10 m
D. 69 m

Answer

To solve the problem of how high an object would travel on the moon when thrown vertically upward with the same initial velocity as on Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Maximum height on Earth, \( H_E = 23.0 \, \text{m} \)
- Acceleration due to gravity on Earth, \( g_E \)
- Acceleration due to gravity on the Moon, \( g_M = \frac{g_E}{6} \)

2. **Use the Kinematic Equation:**
- The kinematic equation that relates initial velocity, final velocity, acceleration, and displacement is:
\[ v^2 = u^2 - 2gH \]
- At the maximum height, the final velocity \( v = 0 \). Therefore, we can rewrite the equation for Earth:
\[ 0 = u^2 - 2g_E H_E \]
- Rearranging gives:
\[ u^2 = 2g_E H_E \]

3. **Calculate Maximum Height on the Moon:**
- For the Moon, we can use the same initial velocity \( u \):
\[ H_M = \frac{u^2}{2g_M} \]
- Substitute \( g_M = \frac{g_E}{6} \) into the equation:
\[ H_M = \frac{u^2}{2 \cdot \frac{g_E}{6}} = \frac{u^2 \cdot 6}{2g_E} = 3 \cdot \frac{u^2}{2g_E} \]

4. **Substitute the Expression for \( \frac{u^2}{2g_E} \):**
- From the earlier step, we know \( \frac{u^2}{2g_E} = H_E \):
\[ H_M = 3 \cdot H_E \]
- Now substitute \( H_E = 23.0 \, \text{m} \):
\[ H_M = 3 \cdot 23.0 \, \text{m} = 69.0 \, \text{m} \]

5. **Final Calculation:**
- Therefore, the maximum height the object would travel on the Moon is:
\[ H_M = 69.0 \, \text{m} \]

### Conclusion:
The object would travel a maximum height of **69.0 meters** on the Moon.