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

Explanation

To solve the problem, we need to find the angle of projection \(\theta\) when the horizontal range \(R\) and the maximum height \(H\) of a projectile are equal.

### Step-by-Step Solution:

1. **Understanding the Relationships**:
The horizontal range \(R\) and the maximum height \(H\) of a projectile launched at an angle \(\theta\) can be expressed as:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]
where \(u\) is the initial velocity and \(g\) is the acceleration due to gravity.

2. **Setting the Condition**:
According to the problem, we have:
\[ R = H \]
Therefore, 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. **Simplifying the Equation**:
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 us:
\[ 2 \sin(\theta) \cos(\theta) = \frac{1}{2} \sin^2(\theta) \]

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:
- Case 1: \(\sin(\theta) = 0\) (which gives \(\theta = 0^\circ\) or \(180^\circ\), not valid for projectile motion)
- Case 2: \(\sin(\theta) - 4 \cos(\theta) = 0\)

7. **Solving for \(\theta\)**:
From Case 2, we have:
\[ \sin(\theta) = 4 \cos(\theta) \]
Dividing both sides by \(\cos(\theta)\) (assuming \(\cos(\theta) \neq 0\)):
\[ \tan(\theta) = 4 \]

8. **Finding the Angle**:
Therefore, we can find \(\theta\) using the inverse tangent function:
\[ \theta = \tan^{-1}(4) \]

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