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 determine the angle of elevation required for a projectile to hit a target on a horizontal plane, given two different scenarios with specified angles and distances. Here’s the step-by-step solution:

### Step 1: Understand the Range Formula
The range \( R \) of a projectile launched with an initial velocity \( u \) at an angle \( \theta \) is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity.

### Step 2: Set Up the Equations for Each Scenario
1. **For the first scenario** (30° elevation):
- The projectile falls 6 m short of the target:
\[ R_1 = R_0 - 6 = \frac{u^2 \sin(60°)}{g} \]
where \( R_0 \) is the range to the target.

2. **For the second scenario** (45° elevation):
- The projectile overshoots the target by 9 m:
\[ R_2 = R_0 + 9 = \frac{u^2 \sin(90°)}{g} \]

### Step 3: Substitute Known Values
- We know that \( \sin(60°) = \frac{\sqrt{3}}{2} \) and \( \sin(90°) = 1 \).
- Substitute these values into the equations:
- For the first scenario:
\[ R_0 - 6 = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} \]
- For the second scenario:
\[ R_0 + 9 = \frac{u^2}{g} \]

### Step 4: Express Both Equations in Terms of \( R_0 \)
1. Rearranging the first equation:
\[ R_0 = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} + 6 \]
2. Rearranging the second equation:
\[ R_0 = \frac{u^2}{g} - 9 \]

### Step 5: Set the Two Expressions for \( R_0 \) Equal
Now we can set the two expressions for \( R_0 \) equal to each other:
\[ \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} + 6 = \frac{u^2}{g} - 9 \]

### Step 6: Solve for \( u^2 \)
1. Multiply through by \( g \) to eliminate the denominator:
\[ u^2 \cdot \frac{\sqrt{3}}{2} + 6g = u^2 - 9g \]
2. Rearranging gives:
\[ u^2 - u^2 \cdot \frac{\sqrt{3}}{2} = 15g \]
3. Factor out \( u^2 \):
\[ u^2 \left(1 - \frac{\sqrt{3}}{2}\right) = 15g \]
4. Solve for \( u^2 \):
\[ u^2 = \frac{15g}{1 - \frac{\sqrt{3}}{2}} = \frac{30g}{2 - \sqrt{3}} \]

### Step 7: Substitute \( u^2 \) into One of the Range Equations
Using the second range equation:
\[ R_0 + 9 = \frac{u^2}{g} \]
Substituting \( u^2 \):
\[ R_0 + 9 = \frac{30g}{g(2 - \sqrt{3})} = \frac{30}{2 - \sqrt{3}} \]
Thus:
\[ R_0 = \frac{30}{2 - \sqrt{3}} - 9 \]

### Step 8: Find \( \sin(2\theta) \)
Using the first range equation:
\[ R_0 - 6 = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} \]
Substituting \( u^2 \):
\[ R_0 - 6 = \frac{15}{2 - \sqrt{3}} \cdot \frac{\sqrt{3}}{2} \]
Now we can find \( \sin(2\theta) \) and subsequently \( \theta \).

### Step 9: Calculate the Angle of Elevation
Finally, we can find \( \theta \) using:
\[ \sin(2\theta) = \frac{R_0}{R_0 + 9} \]
And then take the inverse sine to find \( \theta \).