The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

Options

A. `theta = tan^(-1) ((1)/(4))`
B. `theta = tan^(-1)(4)`
C. `theta = tan^(-1)(2)`
D. `theta = 45^(@)`

Explanation

To solve the problem where the horizontal range and the maximum height of a projectile are equal, we need to find the angle of projection (θ). Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understand the Formulas**:
- 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. **Set the Equations Equal**:
Since the problem states that the range and the maximum height are equal, we can set the two formulas equal to each other:
\[ \frac{u^2 \sin^2(\theta)}{2g} = \frac{u^2 \sin(2\theta)}{g} \]

3. **Cancel Common Terms**:
We can simplify the equation by canceling \(u^2\) and \(g\) from both sides (assuming \(u \neq 0\) and \(g \neq 0\)):
\[ \frac{\sin^2(\theta)}{2} = \sin(2\theta) \]

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

5. **Rearrange the Equation**:
Multiply both sides by 2 to eliminate the fraction:
\[ \sin^2(\theta) = 4 \sin(\theta) \cos(\theta) \]

6. **Factor Out \(\sin(\theta)\)**:
Rearranging gives:
\[ \sin^2(\theta) - 4 \sin(\theta) \cos(\theta) = 0 \]
Factor out \(\sin(\theta)\):
\[ \sin(\theta)(\sin(\theta) - 4 \cos(\theta)) = 0 \]

7. **Solve for \(\theta\)**:
This gives us two cases:
- Case 1: \(\sin(\theta) = 0\) → This implies \(\theta = 0^\circ\) (not a valid angle for projectile motion).
- Case 2: \(\sin(\theta) - 4 \cos(\theta) = 0\) → Rearranging gives:
\[ \sin(\theta) = 4 \cos(\theta) \]
Dividing both sides by \(\cos(\theta)\) (assuming \(\cos(\theta) \neq 0\)):
\[ \tan(\theta) = 4 \]

8. **Find the Angle**:
Therefore, the angle of projection is:
\[ \theta = \tan^{-1}(4) \]

### Final Answer:
The angle of projection of the projectile is \(\theta = \tan^{-1}(4)\).