A rock is thrown downward from an unknown height above the ground with an initial speed of 10 m/s. It strikes the ground 3.0 s later. Determine the initial height of the rock above the ground .

Answer

To determine the initial height of the rock above the ground, we can use the equations of motion. Here's the step-by-step solution:

### Step 1: Identify the known values
- Initial speed (u) = 10 m/s (downward)
- Time (t) = 3 s
- Acceleration (a) = g = 9.8 m/s² (downward)

### Step 2: Set up the equation of motion
We will use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) is the displacement (which will be the height \( h \) but taken as negative since it's downward),
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( t \) is the time.

### Step 3: Substitute the known values into the equation
Since the rock is thrown downward, we take downward as negative:
- \( u = -10 \, \text{m/s} \)
- \( a = -9.8 \, \text{m/s}^2 \)
- \( t = 3 \, \text{s} \)

Substituting these values into the equation:
\[ -h = (-10)(3) + \frac{1}{2}(-9.8)(3^2) \]

### Step 4: Calculate the displacement
Calculating each term:
1. \( -10 \times 3 = -30 \)
2. \( \frac{1}{2} \times -9.8 \times 9 = -44.1 \)

Now, substituting these values back into the equation:
\[ -h = -30 - 44.1 \]
\[ -h = -74.1 \]

### Step 5: Solve for height
To find \( h \):
\[ h = 74.1 \, \text{m} \]

### Final Answer
The initial height of the rock above the ground is **74.1 meters**.

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