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 did it take the object to fall the 8.0 m mentioned ?

Options

A. 0.40 s
B. 1.3 s
C. 0.80 s
D. 2.5 s

Answer

To solve the problem, we will use the equations of motion. Here’s a step-by-step breakdown:

### Step 1: Identify the known values
- Initial velocity (u) = 0 m/s (since the rock is dropped from rest)
- Final velocity (v) = 20.0 m/s
- Distance fallen (s) = 8.0 m

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

Substituting the known values:
\[ (20.0)^2 = (0)^2 + 2a(8.0) \]
\[ 400 = 0 + 16a \]
\[ 16a = 400 \]
\[ a = \frac{400}{16} = 25.0 \, \text{m/s}^2 \]

### Step 3: Use the acceleration to find the time (t)
Now we will use another equation of motion:
\[ v = u + at \]

Substituting the known values:
\[ 20.0 = 0 + (25.0)t \]
\[ 20.0 = 25.0t \]
\[ t = \frac{20.0}{25.0} = 0.8 \, \text{s} \]

### Final Answer
The time taken for the rock to fall 8.0 m is **0.8 seconds**.

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