A particle aimed at a target projected with an angle `15^(@)` with the horizontal is short of the target by 10m .If projected with an angle of `45^(@)` is away from the target by 15m then the angle of projection to hit the target is

Options

A. `1/2sin ^(-1)(1/10)`
B. `1/2sin ^(-1)(3/10)`
C. `1/2sin ^(-1)(9/10)`
D. `1/2sin ^(-1)(7/10)`

Answer

To solve the problem step-by-step, we will use the range formula for projectile motion and set up equations based on the given conditions.

### Step 1: Understand the Problem
We have two scenarios for projectile motion:
1. A particle projected at an angle of \(15^\circ\) falls short of the target by \(10 \, m\).
2. A particle projected at an angle of \(45^\circ\) is \(15 \, m\) beyond the target.

Let \(R\) be the actual range needed to hit the target.

### Step 2: Set Up the Equations
From the first scenario:
- The range when projected at \(15^\circ\) is \(R - 10\).
- The range formula is given by:
\[ R_1 = \frac{u^2 \sin(2\theta_1)}{g} \]
where \(\theta_1 = 15^\circ\).

Thus, we can write:
\[ R - 10 = \frac{u^2 \sin(30^\circ)}{g} \]
Since \(\sin(30^\circ) = \frac{1}{2}\), we have:
\[ R - 10 = \frac{u^2}{2g} \quad \text{(1)} \]

From the second scenario:
- The range when projected at \(45^\circ\) is \(R + 15\).
- The range formula is:
\[ R_2 = \frac{u^2 \sin(2\theta_2)}{g} \]
where \(\theta_2 = 45^\circ\).

Thus, we can write:
\[ R + 15 = \frac{u^2 \sin(90^\circ)}{g} \]
Since \(\sin(90^\circ) = 1\), we have:
\[ R + 15 = \frac{u^2}{g} \quad \text{(2)} \]

### Step 3: Solve the Equations
Now we have two equations:
1. \(R - 10 = \frac{u^2}{2g}\)
2. \(R + 15 = \frac{u^2}{g}\)

From equation (1):
\[ R = \frac{u^2}{2g} + 10 \quad \text{(3)} \]

Substituting equation (3) into equation (2):
\[ \frac{u^2}{2g} + 10 + 15 = \frac{u^2}{g} \]
This simplifies to:
\[ \frac{u^2}{2g} + 25 = \frac{u^2}{g} \]
Subtract \(\frac{u^2}{2g}\) from both sides:
\[ 25 = \frac{u^2}{g} - \frac{u^2}{2g} \]
\[ 25 = \frac{u^2}{2g} \]
Multiplying both sides by \(2g\):
\[ 50g = u^2 \quad \text{(4)} \]

### Step 4: Find the Range \(R\)
Substituting \(u^2\) from equation (4) into equation (3):
\[ R = \frac{50g}{2g} + 10 \]
\[ R = 25 + 10 = 35 \, m \quad \text{(5)} \]

### Step 5: Find the Angle of Projection to Hit the Target
Now we can use the range formula again to find the angle \(\theta\) that would hit the target:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Substituting \(R\) and \(u^2\):
\[ 35 = \frac{50g \sin(2\theta)}{g} \]
This simplifies to:
\[ 35 = 50 \sin(2\theta) \]
\[ \sin(2\theta) = \frac{35}{50} = \frac{7}{10} \]

### Step 6: Solve for \(\theta\)
Taking the inverse sine:
\[ 2\theta = \sin^{-1}\left(\frac{7}{10}\right) \]
Thus,
\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{7}{10}\right) \]

### Final Answer
The angle of projection to hit the target is:
\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{7}{10}\right) \]