Aparticle is projected from the surface of the earth with an intial speed of `4.0 km//s`.Find the maximum height attained by the particle. Radius of earth = 6400km `g = 9.8 m//s^(2)`.

Explanation

To find the maximum height attained by a particle projected from the surface of the Earth, we can use the following formula:

\[ h = \frac{v^2}{2g} - \frac{v^2}{R} \]

where:
- \( h \) is the maximum height,
- \( v \) is the initial speed of the particle,
- \( g \) is the acceleration due to gravity,
- \( R \) is the radius of the Earth.

### Step-by-Step Solution:

1. **Convert the initial speed to SI units**:
The initial speed \( v = 4.0 \, \text{km/s} \) needs to be converted to meters per second (m/s).
\[ v = 4.0 \, \text{km/s} = 4.0 \times 10^3 \, \text{m/s} \]

2. **Identify the values**:
- \( g = 9.8 \, \text{m/s}^2 \)
- \( R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} = 6.4 \times 10^6 \, \text{m} \)

3. **Substitute the values into the formula**:
\[ h = \frac{(4.0 \times 10^3)^2}{2 \times 9.8} - \frac{(4.0 \times 10^3)^2}{6.4 \times 10^6} \]

4. **Calculate \( \frac{(4.0 \times 10^3)^2}{2 \times 9.8} \)**:
\[ \frac{(4.0 \times 10^3)^2}{2 \times 9.8} = \frac{16 \times 10^6}{19.6} \approx 816326.53 \, \text{m} \]

5. **Calculate \( \frac{(4.0 \times 10^3)^2}{6.4 \times 10^6} \)**:
\[ \frac{(4.0 \times 10^3)^2}{6.4 \times 10^6} = \frac{16 \times 10^6}{6.4 \times 10^6} = 2.5 \, \text{m} \]

6. **Combine the results to find \( h \)**:
\[ h = 816326.53 - 2.5 \approx 816324.03 \, \text{m} \]

7. **Convert \( h \) to kilometers**:
\[ h \approx 816.32 \, \text{km} \]

### Final Answer:
The maximum height attained by the particle is approximately \( 816.32 \, \text{km} \).