A projectile is projected from ground such that the maximum height attained by, it is equal to half the horizontal range.The angle of projection with horizontal would be

Options

A. `tan^(-1) (2)`
B. `tan^(-1) (3)`
C. `tan^(-1) (4)`
D. `tan^(-1) sqrt(2)`

Answer

To solve the problem, we need to find the angle of projection (θ) when the maximum height (H) attained by a projectile is equal to half of its horizontal range (R).

### Step-by-Step Solution:

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

2. **Setting Up the Equation**:
- According to the problem, the maximum height is equal to half the horizontal range:
\[ H = \frac{1}{2} R \]
- Substituting the formulas for H and R into this equation gives:
\[ \frac{u^2 \sin^2 \theta}{2g} = \frac{1}{2} \left( \frac{2u^2 \sin \theta \cos \theta}{g} \right) \]

3. **Simplifying the Equation**:
- The right side simplifies to:
\[ \frac{u^2 \sin \theta \cos \theta}{g} \]
- Now we have:
\[ \frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin \theta \cos \theta}{g} \]
- We can cancel \(u^2\) and \(g\) from both sides (assuming \(u \neq 0\) and \(g \neq 0\)):
\[ \frac{\sin^2 \theta}{2} = \sin \theta \cos \theta \]

4. **Rearranging the Equation**:
- Multiply both sides by 2:
\[ \sin^2 \theta = 2 \sin \theta \cos \theta \]
- This can be rewritten using the identity \( \sin(2\theta) = 2 \sin \theta \cos \theta \):
\[ \sin^2 \theta = \sin(2\theta) \]

5. **Dividing by \(\sin \theta\)** (assuming \(\sin \theta \neq 0\)):
- We can divide both sides by \(\sin \theta\):
\[ \sin \theta = 2 \cos \theta \]
- This can be rewritten as:
\[ \tan \theta = 2 \]

6. **Finding the Angle**:
- Therefore, the angle of projection θ can be found using:
\[ \theta = \tan^{-1}(2) \]

### Final Answer:
The angle of projection with the horizontal is:
\[ \theta = \tan^{-1}(2) \]