At what angle to the horizontal should an object be projected so that the maximum height reached is equal to the horizontal range.

Options

A. `tan theta = 2`
B. `tan theta = 4`
C. `tan theta = 3`
D. `tan theta = 3`

Explanation

To solve the problem of finding the angle at which an object should be projected so that the maximum height reached is equal to the horizontal range, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formulas**:
- The formula for maximum height \( h_{max} \) reached by a projectile is given by:
\[ h_{max} = \frac{u^2 \sin^2 \theta}{2g} \]
- The formula for the horizontal range \( R \) of a projectile is:
\[ R = \frac{u^2 \sin 2\theta}{g} \]

2. **Set the Condition**:
- According to the problem, we need to set the maximum height equal to the horizontal range:
\[ h_{max} = R \]

3. **Substitute the Formulas**:
- Substitute the formulas for \( h_{max} \) and \( R \) into the equation:
\[ \frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin 2\theta}{g} \]

4. **Simplify the Equation**:
- Cancel \( u^2 \) and \( g \) from both sides (assuming \( u \neq 0 \) and \( g \neq 0 \)):
\[ \frac{\sin^2 \theta}{2} = \sin 2\theta \]
- Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ \frac{\sin^2 \theta}{2} = 2 \sin \theta \cos \theta \]

5. **Rearrange the Equation**:
- Multiply both sides by 2:
\[ \sin^2 \theta = 4 \sin \theta \cos \theta \]
- 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. \( \sin \theta - 4 \cos \theta = 0 \)

7. **Solve for \( \theta \)**:
- From \( \sin \theta = 4 \cos \theta \), we can write:
\[ \tan \theta = 4 \]
- Therefore, the angle \( \theta \) can be found using:
\[ \theta = \tan^{-1}(4) \]

8. **Conclusion**:
- The angle at which the object should be projected is:
\[ \theta = \tan^{-1}(4) \]