A body is projected with velocity `u` such that in horizontal range and maximum vertical heights are same.The maximum height is

Options

A. `(8u^2)/17g`
B. `(u^2)/(2g)`
C. `(16u^2)/(7g)`
D. `(4u^2)/(3g)`

Explanation

To solve the problem, we need to find the maximum vertical height of a body projected with velocity \( u \) such that the horizontal range and maximum vertical height are equal.

### Step-by-Step Solution:

1. **Understanding the Equations**:
- The formula for the horizontal range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
- The formula for the maximum height \( H \) of a projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

2. **Setting the Equations Equal**:
- According to the problem, the horizontal range and maximum height are equal:
\[ R = H \]
- Therefore, we can write:
\[ \frac{u^2 \sin 2\theta}{g} = \frac{u^2 \sin^2 \theta}{2g} \]

3. **Simplifying the Equation**:
- Cancel \( u^2 \) and \( g \) from both sides (assuming \( u \neq 0 \) and \( g \neq 0 \)):
\[ \sin 2\theta = \frac{1}{2} \sin^2 \theta \]
- Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), we can rewrite the equation:
\[ 2 \sin \theta \cos \theta = \frac{1}{2} \sin^2 \theta \]

4. **Rearranging the Equation**:
- Multiply both sides by 2 to eliminate the fraction:
\[ 4 \sin \theta \cos \theta = \sin^2 \theta \]
- Rearranging gives:
\[ \sin^2 \theta - 4 \sin \theta \cos \theta = 0 \]
- Factoring out \( \sin \theta \):
\[ \sin \theta (\sin \theta - 4 \cos \theta) = 0 \]

5. **Finding \( \tan \theta \)**:
- Since \( \sin \theta \neq 0 \) for a projectile, we have:
\[ \sin \theta = 4 \cos \theta \]
- Dividing both sides by \( \cos \theta \) gives:
\[ \tan \theta = 4 \]

6. **Calculating Maximum Height**:
- We can find \( \sin^2 \theta \) using \( \tan \theta \):
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \implies \sin \theta = 4 \cos \theta \]
- Using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ (4 \cos \theta)^2 + \cos^2 \theta = 1 \implies 16 \cos^2 \theta + \cos^2 \theta = 1 \]
\[ 17 \cos^2 \theta = 1 \implies \cos^2 \theta = \frac{1}{17} \]
- Thus, \( \sin^2 \theta = 4^2 \cos^2 \theta = 16 \cdot \frac{1}{17} = \frac{16}{17} \).

7. **Substituting to Find Maximum Height**:
- Now substituting \( \sin^2 \theta \) into the height formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \cdot \frac{16}{17}}{2g} = \frac{8u^2}{17g} \]

### Final Answer:
The maximum vertical height \( H \) is:
\[ H = \frac{8u^2}{17g} \]