If R and h represent the horizontal range and maximum height respectively of an obliquie projectile, the `(R^(2))/(8h)+2h` represents

Options

A. maximum horizontal range
B. maximum vertical range
C. time of flight
D. velocity of projectile at highest point

Answer

To solve the problem, we need to analyze the expressions for the horizontal range \( R \) and the maximum height \( h \) of an oblique projectile.

### Step-by-Step Solution:

1. **Define the expressions for Range and Height**:
- The horizontal range \( R \) of a projectile launched with an initial velocity \( u \) at an angle \( \theta \) is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
- The maximum height \( h \) is given by:
\[ h = \frac{u^2 \sin^2(\theta)}{2g} \]

2. **Calculate \( R^2 \) and \( 8h \)**:
- First, we calculate \( R^2 \):
\[ R^2 = \left(\frac{u^2 \sin(2\theta)}{g}\right)^2 = \frac{u^4 \sin^2(2\theta)}{g^2} \]
- Now, calculate \( 8h \):
\[ 8h = 8 \times \frac{u^2 \sin^2(\theta)}{2g} = \frac{4u^2 \sin^2(\theta)}{g} \]

3. **Find \( \frac{R^2}{8h} \)**:
- Now we can find \( \frac{R^2}{8h} \):
\[ \frac{R^2}{8h} = \frac{\frac{u^4 \sin^2(2\theta)}{g^2}}{\frac{4u^2 \sin^2(\theta)}{g}} = \frac{u^4 \sin^2(2\theta)}{g^2} \cdot \frac{g}{4u^2 \sin^2(\theta)} = \frac{u^2 \sin^2(2\theta)}{4g \sin^2(\theta)} \]

4. **Simplify \( \sin(2\theta) \)**:
- Recall that \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \):
\[ \sin^2(2\theta) = 4 \sin^2(\theta) \cos^2(\theta) \]
- Substitute this back into the equation:
\[ \frac{R^2}{8h} = \frac{u^2 \cdot 4 \sin^2(\theta) \cos^2(\theta)}{4g \sin^2(\theta)} = \frac{u^2 \cos^2(\theta)}{g} \]

5. **Combine \( \frac{R^2}{8h} \) and \( 2h \)**:
- Now, calculate \( 2h \):
\[ 2h = 2 \times \frac{u^2 \sin^2(\theta)}{2g} = \frac{u^2 \sin^2(\theta)}{g} \]
- Add \( \frac{R^2}{8h} \) and \( 2h \):
\[ \frac{R^2}{8h} + 2h = \frac{u^2 \cos^2(\theta)}{g} + \frac{u^2 \sin^2(\theta)}{g} = \frac{u^2 (\sin^2(\theta) + \cos^2(\theta))}{g} \]

6. **Use the Pythagorean Identity**:
- Since \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ \frac{R^2}{8h} + 2h = \frac{u^2}{g} \]

7. **Conclusion**:
- Therefore, the expression \( \frac{R^2}{8h} + 2h \) simplifies to:
\[ \frac{u^2}{g} \]
- This represents the maximum range of the projectile when \( \theta \) is optimized.

### Final Answer:
The expression \( \frac{R^2}{8h} + 2h \) represents the maximum range of the projectile, which is \( \frac{u^2}{g} \).