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.
Determine the speed of the object after falling from rest through 16 m on this planet .

Options

A. 28 m/s
B. 32 m/s
C. 56 m/s
D. 64 m/s

Answer

To solve the problem, we will use the equations of motion. Let's break it down step by step.

### Step 1: Identify the given data
- Initial velocity (u) = 0 m/s (the rock is dropped from rest)
- Final velocity (v) after falling 8 m = 20.0 m/s
- Distance fallen (s) = 8.0 m

### Step 2: Use the equation of motion to find acceleration (a)
We will use the equation of motion:
\[ v^2 = u^2 + 2as \]

Substituting the known values:
\[ (20.0)^2 = (0)^2 + 2a(8.0) \]

Calculating:
\[ 400 = 0 + 16a \]
\[ 16a = 400 \]
\[ a = \frac{400}{16} = 25 \, \text{m/s}^2 \]

### Step 3: Determine the speed after falling 16 m
Now we need to find the final velocity (v) after falling a distance of 16 m from rest. We will again use the equation of motion:
\[ v^2 = u^2 + 2as \]

Here, the initial velocity (u) is still 0 m/s, the acceleration (a) is 25 m/s², and the distance (s) is 16 m.

Substituting the values:
\[ v^2 = (0)^2 + 2(25)(16) \]

Calculating:
\[ v^2 = 0 + 50 \times 16 \]
\[ v^2 = 800 \]

### Step 4: Calculate the final velocity (v)
Taking the square root of both sides:
\[ v = \sqrt{800} \]
\[ v = 20\sqrt{2} \, \text{m/s} \]
\[ v \approx 28.28 \, \text{m/s} \]

### Final Answer
The speed of the object after falling from rest through 16 m on this planet is approximately **28.28 m/s**.

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