A spherical ball of radius `3.0xx10^(-4) m` and density `10^(4) kg//m^(3)` falls freely under gravity through a distance h before entering a tank of water. If after entering the water the velocity of the ball does not change, find h. Viscosity of water is `9.8xx10^(6) N-s//m^(2)`.

Explanation

To solve the problem, we need to find the distance \( h \) that the spherical ball falls freely under gravity before entering the tank of water, given that its velocity remains constant after entering the water. This constant velocity is the terminal velocity of the ball in water.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Radius of the ball, \( R = 3.0 \times 10^{-4} \, \text{m} \)
- Density of the ball, \( \rho = 10^4 \, \text{kg/m}^3 \)
- Density of water, \( \sigma = 10^3 \, \text{kg/m}^3 \)
- Viscosity of water, \( \eta = 9.8 \times 10^{-6} \, \text{N s/m}^2 \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

2. **Determine the Velocity Before Entering Water:**
The velocity \( V \) of the ball just before it enters the water can be calculated using the equation for free fall:
\[ V = \sqrt{2gh} \]

3. **Determine the Terminal Velocity in Water:**
The terminal velocity \( V_t \) of the ball when it enters the water can be calculated using Stokes' law:
\[ V_t = \frac{2}{9} \cdot \frac{R^2 (\rho - \sigma) g}{\eta} \]

4. **Set the Velocities Equal:**
Since the velocity before entering the water is equal to the terminal velocity after entering the water, we can set the two equations equal to each other:
\[ \sqrt{2gh} = \frac{2}{9} \cdot \frac{R^2 (\rho - \sigma) g}{\eta} \]

5. **Square Both Sides:**
Squaring both sides to eliminate the square root gives:
\[ 2gh = \left(\frac{2}{9} \cdot \frac{R^2 (\rho - \sigma) g}{\eta}\right)^2 \]

6. **Rearranging for \( h \):**
Rearranging the equation to solve for \( h \):
\[ h = \frac{1}{2g} \left(\frac{2}{9} \cdot \frac{R^2 (\rho - \sigma) g}{\eta}\right)^2 \]

7. **Substituting Values:**
Substitute the known values into the equation:
\[ h = \frac{1}{2 \cdot 9.8} \left(\frac{2}{9} \cdot \frac{(3 \times 10^{-4})^2 (10^4 - 10^3) \cdot 9.8}{9.8 \times 10^{-6}}\right)^2 \]

8. **Calculating \( h \):**
Calculate the value step by step:
- Calculate \( R^2 = (3 \times 10^{-4})^2 = 9 \times 10^{-8} \, \text{m}^2 \)
- Calculate \( \rho - \sigma = 10^4 - 10^3 = 9 \times 10^3 \, \text{kg/m}^3 \)
- Substitute these values back into the equation and simplify.

9. **Final Calculation:**
After performing the calculations, we find:
\[ h \approx 1.65 \times 10^3 \, \text{m} \]

### Final Answer:
The distance \( h \) that the ball falls before entering the water is approximately \( 1.65 \times 10^3 \, \text{m} \).