A particle is projected from the ground with speed u un such a manner that its range is found to be two times the maximum height . If range of the projectile can be written as `("nu"^(2))/(5g)` , then what is the value of n ?

Explanation

To solve the problem, we need to find the value of \( n \) given that the range \( R \) of a projectile is twice the maximum height \( H \).

### Step-by-Step Solution:

1. **Understanding the Relationships**:
- The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
- The maximum height \( H \) of a projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
- According to the problem, the range is twice the maximum height:
\[ R = 2H \]

2. **Substituting the Formulas**:
- Substitute the formulas for \( R \) and \( H \):
\[ \frac{u^2 \sin 2\theta}{g} = 2 \left( \frac{u^2 \sin^2 \theta}{2g} \right) \]

3. **Simplifying the Equation**:
- The \( g \) cancels out from both sides:
\[ u^2 \sin 2\theta = u^2 \sin^2 \theta \]
- Dividing both sides by \( u^2 \) (assuming \( u \neq 0 \)):
\[ \sin 2\theta = \sin^2 \theta \]

4. **Using the Double Angle Identity**:
- Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ 2 \sin \theta \cos \theta = \sin^2 \theta \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ 2 \cos \theta = \sin \theta \]
- Dividing both sides by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)):
\[ 2 = \tan \theta \]

6. **Finding the Angle**:
- From \( \tan \theta = 2 \), we can find \( \theta \):
\[ \theta = \tan^{-1}(2) \]

7. **Finding the Range**:
- Now we need to calculate the range \( R \):
\[ R = \frac{u^2 \sin 2\theta}{g} \]
- Using \( \sin 2\theta = 2 \sin \theta \cos \theta \):
- We know \( \sin \theta = \frac{2}{\sqrt{5}} \) and \( \cos \theta = \frac{1}{\sqrt{5}} \) from the triangle formed by \( \tan \theta = 2 \).
- Therefore:
\[ \sin 2\theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5} \]
- Substitute this back into the range formula:
\[ R = \frac{u^2 \cdot \frac{4}{5}}{g} = \frac{4u^2}{5g} \]

8. **Identifying \( n \)**:
- We are given that \( R = \frac{nu^2}{5g} \).
- Comparing both expressions for \( R \):
\[ \frac{4u^2}{5g} = \frac{nu^2}{5g} \]
- This implies:
\[ n = 4 \]

### Final Answer:
The value of \( n \) is \( 4 \).