If the water falls from a dam into a turbine wheel 19.6 m below, then the velocity of water at the turbine is `(g=9.8m//s^(2))`

Options

A. a.9.8 m/s
B. b.19.6 m/s
C. c.39.2 m/s
D. d.98.0 m/s

Explanation

To find the velocity of water at the turbine when it falls from a height of 19.6 meters, we can use the principle of conservation of energy. The potential energy of the water at the height will convert into kinetic energy at the turbine.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Height (h) = 19.6 m
- Acceleration due to gravity (g) = 9.8 m/s²

2. **Write the Energy Conservation Equation:**
The potential energy (PE) at the height will equal the kinetic energy (KE) at the turbine:
\[ PE = KE \]
\[ mgh = \frac{1}{2} mv^2 \]
Here, \(m\) is the mass of the water, \(g\) is the acceleration due to gravity, \(h\) is the height, and \(v\) is the final velocity.

3. **Cancel the Mass (m):**
Since mass appears on both sides of the equation, we can cancel it out:
\[ gh = \frac{1}{2} v^2 \]

4. **Rearrange the Equation to Solve for Velocity (v):**
Multiply both sides by 2:
\[ 2gh = v^2 \]
Now take the square root of both sides:
\[ v = \sqrt{2gh} \]

5. **Substitute the Values:**
Substitute \(g = 9.8 \, \text{m/s}^2\) and \(h = 19.6 \, \text{m}\):
\[ v = \sqrt{2 \times 9.8 \times 19.6} \]

6. **Calculate the Value:**
First, calculate \(2 \times 9.8 \times 19.6\):
\[ 2 \times 9.8 = 19.6 \]
\[ 19.6 \times 19.6 = 384.16 \]
Now take the square root:
\[ v = \sqrt{384.16} \approx 19.6 \, \text{m/s} \]

### Final Answer:
The velocity of the water at the turbine is approximately **19.6 m/s**.

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