The angle of projection at which the horizontal range and maximum height of projectile are equal is

Options

A. `45^@`
B. `theta = tan^(-1) (0.25)`
C. `theta = tan^(-1) (4)`
D. `60^(@)`

Answer

To solve the problem of finding the angle of projection at which the horizontal range and maximum height of a projectile are equal, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formulas**:
- The horizontal range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
- The maximum height \( H \) of a projectile is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

2. **Set the Range Equal to the Height**:
- According to the problem, we need to find the angle \( \theta \) such that:
\[ R = H \]
- 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. **Cancel Common Terms**:
- Since \( u^2 \) and \( g \) are common on both sides, we can cancel them out:
\[ \sin 2\theta = \frac{1}{2} \sin^2 \theta \]

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

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. **Factor the Equation**:
- Factor out \( \sin \theta \):
\[ \sin \theta (\sin \theta - 4 \cos \theta) = 0 \]
- This gives us two cases:
1. \( \sin \theta = 0 \) (which is not a valid angle for projection)
2. \( \sin \theta - 4 \cos \theta = 0 \) or \( \sin \theta = 4 \cos \theta \)

7. **Divide by Cosine**:
- Dividing both sides by \( \cos \theta \) gives:
\[ \tan \theta = 4 \]

8. **Find the Angle**:
- To find \( \theta \), take the inverse tangent:
\[ \theta = \tan^{-1}(4) \]

### Final Answer:
The angle of projection at which the horizontal range and maximum height of the projectile are equal is:
\[ \theta = \tan^{-1}(4) \]