If R and H represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal is

Options

A. `tan^(-1)((H)/(R ))`
B. `tan^(-1)((2H)/(R ))`
C. `tan^(-1) ((4H)/(R ))`
D. `tan^(-1) ((4H)/(H))`

Explanation

To find the angle of projection (θ) with respect to the horizontal when given the horizontal range (R) and maximum height (H) of a projectile, we can follow these steps:

### Step 1: Write the formulas for range and maximum height
The formulas for the horizontal range (R) and maximum height (H) of a projectile are given by:

- \( R = \frac{u^2 \sin 2\theta}{g} \) (1)
- \( H = \frac{u^2 \sin^2 \theta}{2g} \) (2)

Where:
- \( u \) is the initial velocity,
- \( g \) is the acceleration due to gravity,
- \( \theta \) is the angle of projection.

### Step 2: Find the ratio of height to range
To find the angle of projection, we can take the ratio of maximum height (H) to horizontal range (R):

\[ \frac{H}{R} = \frac{\frac{u^2 \sin^2 \theta}{2g}}{\frac{u^2 \sin 2\theta}{g}} \]

### Step 3: Simplify the ratio
Now, simplify the expression:

\[ \frac{H}{R} = \frac{u^2 \sin^2 \theta}{2g} \cdot \frac{g}{u^2 \sin 2\theta} \]

This simplifies to:

\[ \frac{H}{R} = \frac{\sin^2 \theta}{2 \sin 2\theta} \]

### Step 4: Substitute \( \sin 2\theta \)
Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Therefore, we can substitute this into our equation:

\[ \frac{H}{R} = \frac{\sin^2 \theta}{2 \cdot 2 \sin \theta \cos \theta} = \frac{\sin^2 \theta}{4 \sin \theta \cos \theta} \]

### Step 5: Cancel out terms
Now, we can cancel one \( \sin \theta \) from the numerator and denominator:

\[ \frac{H}{R} = \frac{\sin \theta}{4 \cos \theta} \]

### Step 6: Express in terms of \( \tan \theta \)
This can be rewritten as:

\[ \frac{H}{R} = \frac{1}{4} \tan \theta \]

### Step 7: Solve for \( \tan \theta \)
Rearranging gives us:

\[ \tan \theta = \frac{4H}{R} \]

### Step 8: Find \( \theta \)
Finally, to find the angle of projection \( \theta \):

\[ \theta = \tan^{-1} \left( \frac{4H}{R} \right) \]

### Conclusion
Thus, the angle of projection with the horizontal is given by:

\[ \theta = \tan^{-1} \left( \frac{4H}{R} \right) \]