A projectile is aimed at a mark on a horizontal plane through the point of projection and falls 6 m short when its elevation is 30° but overshoots the mark by 9 m when its elevation is `45^@`. The angle of elevation of projectile to hit the target on the horizontal plane is :

Options

A. `sin^(-1)[(3sqrt(3)+4)/5]`
B. `cos^(-1)[(3sqrt(3)+4)/5]`
C. `1/2cos^(-1)[(3sqrt(3)+4)/10]`
D. `1/2sin^(-1)[(3sqrt(3)+4)/10]`

Explanation

To solve the problem, we need to find the angle of elevation (θ) for a projectile that will hit a target on a horizontal plane. We know the projectile falls short by 6 m when fired at 30° and overshoots by 9 m when fired at 45°.

### Step-by-Step Solution:

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

2. **Setting Up the Equations**:
- For \( \theta = 30° \):
The projectile falls 6 m short, so the range can be expressed as:
\[ R_{30} = \frac{u^2 \sin(60°)}{g} = R - 6 \]
where \( R \) is the actual distance to the target.
Thus, we have:
\[ \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} = R - 6 \quad \text{(Equation 1)} \]

- For \( \theta = 45° \):
The projectile overshoots by 9 m, so:
\[ R_{45} = \frac{u^2 \sin(90°)}{g} = R + 9 \]
Thus, we have:
\[ \frac{u^2}{g} = R + 9 \quad \text{(Equation 2)} \]

3. **Eliminating R**:
From Equation 1:
\[ R = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} + 6 \]
Substitute this expression for \( R \) into Equation 2:
\[ \frac{u^2}{g} = \left(\frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} + 6\right) + 9 \]
Simplifying gives:
\[ \frac{u^2}{g} = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} + 15 \]

4. **Rearranging the Equation**:
Multiply through by \( g \) to eliminate the denominator:
\[ u^2 = u^2 \cdot \frac{\sqrt{3}}{2} + 15g \]
Rearranging gives:
\[ u^2 - u^2 \cdot \frac{\sqrt{3}}{2} = 15g \]
Factoring out \( u^2 \):
\[ u^2 \left(1 - \frac{\sqrt{3}}{2}\right) = 15g \]
Thus,
\[ u^2 = \frac{15g}{1 - \frac{\sqrt{3}}{2}} \]

5. **Finding sin(2θ)**:
Substitute \( u^2 \) back into either Equation 1 or Equation 2 to find \( \sin(2\theta) \):
Using Equation 1:
\[ R - 6 = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} \]
Substitute \( u^2 \):
\[ R - 6 = \frac{\frac{15g}{1 - \frac{\sqrt{3}}{2}} \cdot \frac{\sqrt{3}}{2}}{g} \]
Simplifying gives:
\[ R - 6 = \frac{15\sqrt{3}/2}{1 - \frac{\sqrt{3}}{2}} \]

6. **Finding the angle θ**:
Rearranging gives us:
\[ \sin(2\theta) = \frac{R + 9}{u^2/g} \]
Substitute \( u^2 \) to find \( \theta \):
\[ 2\theta = \sin^{-1}\left(\frac{4 + 3\sqrt{3}}{10}\right) \]
Finally, solve for \( \theta \):
\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{4 + 3\sqrt{3}}{10}\right) \]

### Final Answer:
The angle of elevation \( \theta \) to hit the target is:
\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{4 + 3\sqrt{3}}{10}\right) \]