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

Options

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

Answer

To solve the problem, we need to find the maximum height \( H \) of a projectile when the horizontal range and the maximum height are equal, given the initial velocity \( u \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that the horizontal range \( R \) and the maximum height \( H \) of a projectile are given by the formulas:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
According to the problem, \( R = H \).

2. **Setting the Equations Equal**:
Since \( R = H \), we can set the two equations equal to each other:
\[ \frac{u^2 \sin 2\theta}{g} = \frac{u^2 \sin^2 \theta}{2g} \]

3. **Canceling Common Terms**:
We can cancel \( u^2 \) and \( g \) from both sides (assuming \( u \neq 0 \) and \( g \neq 0 \)):
\[ \sin 2\theta = \frac{1}{2} \sin^2 \theta \]

4. **Using the Double Angle Identity**:
Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Substituting this into the equation gives:
\[ 2 \sin \theta \cos \theta = \frac{1}{2} \sin^2 \theta \]

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

6. **Factoring the Equation**:
Factor out \( \sin \theta \):
\[ \sin \theta (\sin \theta - 4 \cos \theta) = 0 \]
This gives us two cases: \( \sin \theta = 0 \) (which is not valid for projectile motion) or \( \sin \theta = 4 \cos \theta \).

7. **Finding the Value of \( \tan \theta \)**:
From \( \sin \theta = 4 \cos \theta \), we can write:
\[ \tan \theta = 4 \]

8. **Using the Pythagorean Identity**:
Using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \):
\[ 1 + 16 = \sec^2 \theta \implies \sec^2 \theta = 17 \implies \cos^2 \theta = \frac{1}{17} \]
Therefore,
\[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{1}{17} = \frac{16}{17} \]

9. **Calculating Maximum Height**:
Now, substitute \( \sin^2 \theta \) back into the formula for maximum height \( H \):
\[ H = \frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \cdot \frac{16}{17}}{2g} = \frac{8u^2}{17g} \]

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