Calculate the height at which a man's weight becomes `( 4//9)` of his weight on the surface of the earth, if the radius of the earth is 6400km.

Explanation

To solve the problem of calculating the height at which a man's weight becomes \( \frac{4}{9} \) of his weight on the surface of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between weight and height**:
The weight of an object at a height \( h \) above the Earth's surface is given by the formula:
\[ W_h = W_s \cdot \frac{R^2}{(R + h)^2} \]
where:
- \( W_h \) is the weight at height \( h \),
- \( W_s \) is the weight on the surface of the Earth,
- \( R \) is the radius of the Earth.

2. **Set up the equation**:
According to the problem, the weight at height \( h \) is \( \frac{4}{9} \) of the weight on the surface:
\[ W_h = \frac{4}{9} W_s \]
Substituting the expression for \( W_h \):
\[ W_s \cdot \frac{R^2}{(R + h)^2} = \frac{4}{9} W_s \]

3. **Cancel \( W_s \)** (assuming \( W_s \neq 0 \)):
\[ \frac{R^2}{(R + h)^2} = \frac{4}{9} \]

4. **Cross-multiply to eliminate the fraction**:
\[ 9R^2 = 4(R + h)^2 \]

5. **Expand the right side**:
\[ 9R^2 = 4(R^2 + 2Rh + h^2) \]
\[ 9R^2 = 4R^2 + 8Rh + 4h^2 \]

6. **Rearrange the equation**:
\[ 9R^2 - 4R^2 = 8Rh + 4h^2 \]
\[ 5R^2 = 8Rh + 4h^2 \]

7. **Rearrange to form a quadratic equation**:
\[ 4h^2 + 8Rh - 5R^2 = 0 \]

8. **Use the quadratic formula**:
The quadratic formula is given by:
\[ h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 4 \), \( b = 8R \), and \( c = -5R^2 \):
\[ h = \frac{-8R \pm \sqrt{(8R)^2 - 4 \cdot 4 \cdot (-5R^2)}}{2 \cdot 4} \]
\[ h = \frac{-8R \pm \sqrt{64R^2 + 80R^2}}{8} \]
\[ h = \frac{-8R \pm \sqrt{144R^2}}{8} \]
\[ h = \frac{-8R \pm 12R}{8} \]

9. **Calculate the two possible values for \( h \)**:
- Taking the positive root:
\[ h = \frac{4R}{8} = \frac{R}{2} \]

10. **Substitute the value of \( R \)**:
Given \( R = 6400 \) km:
\[ h = \frac{6400 \text{ km}}{2} = 3200 \text{ km} \]

### Final Answer:
The height at which a man's weight becomes \( \frac{4}{9} \) of his weight on the surface of the Earth is **3200 km**.