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.
How long would it take the object, falling from rest, to fall 16 m on this planet ?

Options

A. 0.8 s
B. 2.5 s
C. 1.1 s
D. 3.5 s

Answer

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

### Step 1: Identify the known values
- Initial velocity \( u = 0 \) m/s (the rock is dropped from rest)
- Final velocity after falling 8 m, \( v = 20 \) m/s
- Distance fallen, \( s = 8 \) m

### Step 2: Use the equation of motion to find acceleration
We can use the equation of motion:
\[ v^2 = u^2 + 2as \]
Substituting the known values:
\[ 20^2 = 0^2 + 2a(8) \]
This simplifies to:
\[ 400 = 16a \]
Now, solve for \( a \):
\[ a = \frac{400}{16} = 25 \, \text{m/s}^2 \]

### Step 3: Calculate the time to fall 16 m
Now we need to find the time \( t \) it takes to fall 16 m. We will use the equation:
\[ s = ut + \frac{1}{2}at^2 \]
Substituting the known values:
\[ 16 = 0 \cdot t + \frac{1}{2}(25)t^2 \]
This simplifies to:
\[ 16 = \frac{25}{2}t^2 \]
Multiply both sides by 2:
\[ 32 = 25t^2 \]
Now, solve for \( t^2 \):
\[ t^2 = \frac{32}{25} \]
Taking the square root of both sides gives:
\[ t = \sqrt{\frac{32}{25}} = \frac{\sqrt{32}}{5} = \frac{4\sqrt{2}}{5} \approx 1.13 \, \text{s} \]

### Final Answer
The time it takes for the rock to fall 16 m is approximately **1.13 seconds**.

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