A rock, dropped from rest near the surface of an atmosphere-free planet, attains a speed of 20.0 m/s after falling 8.0 m.
What is the magnitude of the acceleration due to gravity on the surface of this planet ?

Options

A. `0.40 m//s^(2)`
B. `2.5 m//s^(2)`
C. `1.3 m//s^(2)`
D. `25 m//s^(2)`

Answer

To find the magnitude of the acceleration due to gravity on the surface of the planet, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement. The equation we will use is:

\[ v^2 = u^2 + 2as \]

Where:
- \( v \) = final velocity (20.0 m/s)
- \( u \) = initial velocity (0 m/s, since the rock is dropped from rest)
- \( a \) = acceleration (which we are trying to find)
- \( s \) = displacement (8.0 m)

### Step-by-Step Solution:

1. **Identify the known values:**
- Final velocity \( v = 20.0 \, \text{m/s} \)
- Initial velocity \( u = 0 \, \text{m/s} \)
- Displacement \( s = 8.0 \, \text{m} \)

2. **Substitute the known values into the kinematic equation:**
\[ v^2 = u^2 + 2as \]
Plugging in the values:
\[ (20.0)^2 = (0)^2 + 2a(8.0) \]

3. **Calculate \( v^2 \):**
\[ 400 = 0 + 16a \]

4. **Rearrange the equation to solve for \( a \):**
\[ 400 = 16a \]
\[ a = \frac{400}{16} \]

5. **Calculate \( a \):**
\[ a = 25 \, \text{m/s}^2 \]

### Conclusion:
The magnitude of the acceleration due to gravity on the surface of this planet is \( 25 \, \text{m/s}^2 \).