Two guns situated at the top of a hill of height `10 m` fire one shot each with the same speed `5sqrt(3) m//s` at some interval of time. One gun fires horizontal and the other fores upwards at an angle of `60^(@)` with the horizontal. Two shots collide in air at a poit `P`. Find (i) time-interval between the firing and (ii) coordinates of the point `P`. Take the origin of coordinates system at the foot of the hill right below the muzzle and trajectorise in the `x-y` plane.

Explanation

To solve the problem, we need to analyze the motion of both bullets fired from the guns. One gun fires horizontally, while the other fires at an angle of 60 degrees with the horizontal. We will find the time interval between the firings and the coordinates of the collision point.

### Step 1: Determine the initial velocities of both bullets
- The speed of both bullets is given as \( v = 5\sqrt{3} \, \text{m/s} \).
- For the bullet fired horizontally, the initial velocity components are:
- \( v_{1x} = 5\sqrt{3} \, \text{m/s} \) (horizontal)
- \( v_{1y} = 0 \, \text{m/s} \) (vertical)

- For the bullet fired at an angle of 60 degrees:
- \( v_{2x} = v \cos(60^\circ) = 5\sqrt{3} \cdot \frac{1}{2} = \frac{5\sqrt{3}}{2} \, \text{m/s} \)
- \( v_{2y} = v \sin(60^\circ) = 5\sqrt{3} \cdot \frac{\sqrt{3}}{2} = \frac{15}{2} \, \text{m/s} \)

### Step 2: Analyze the motion of the bullets
- The horizontal bullet travels with a constant velocity, while the bullet fired at an angle experiences vertical motion due to gravity.
- The height of the hill is \( h = 10 \, \text{m} \).
- The time taken for the horizontal bullet to fall \( h \) can be calculated using the equation of motion:
\[ h = \frac{1}{2} g t^2 \quad \Rightarrow \quad 10 = \frac{1}{2} \cdot 9.8 \cdot t^2 \quad \Rightarrow \quad t^2 = \frac{20}{9.8} \quad \Rightarrow \quad t \approx 1.43 \, \text{s} \]

### Step 3: Determine the time of flight for the angled bullet
- The time of flight \( t \) for the angled bullet can be found using the vertical motion equation:
\[ y = v_{2y} t - \frac{1}{2} g t^2 \]
Setting \( y = -10 \, \text{m} \) (since it falls down):
\[ -10 = \frac{15}{2} t - \frac{1}{2} \cdot 9.8 t^2 \]

### Step 4: Set up the equations for collision
- Let \( \Delta t \) be the time interval between the firings. The horizontal bullet travels a distance \( x_1 = v_{1x} (t) \) and the angled bullet travels \( x_2 = v_{2x} (t + \Delta t) \).
- For collision, both bullets must cover the same horizontal distance:
\[ v_{1x} t = v_{2x} (t + \Delta t) \]

### Step 5: Solve for the time interval \( \Delta t \)
- Substitute the values:
\[ 5\sqrt{3} t = \frac{5\sqrt{3}}{2} (t + \Delta t) \]
- Simplifying gives:
\[ 10t = 5(t + \Delta t) \quad \Rightarrow \quad 10t = 5t + 5\Delta t \quad \Rightarrow \quad 5t = 5\Delta t \quad \Rightarrow \quad \Delta t = t \]
- Since we calculated \( t \approx 1.43 \, \text{s} \), we find \( \Delta t \approx 1.43 \, \text{s} \).

### Step 6: Find the coordinates of point P
- Calculate the horizontal distance traveled by the horizontal bullet:
\[ x_1 = v_{1x} t = 5\sqrt{3} \cdot 1.43 \approx 12.35 \, \text{m} \]
- For the bullet fired at an angle, calculate its vertical position when it reaches point P:
- Time of flight for the angled bullet is \( t + \Delta t = 2t \).
- Calculate the vertical position:
\[ y = \frac{15}{2} (2t) - \frac{1}{2} \cdot 9.8 (2t)^2 \]
- Substitute \( t \approx 1.43 \):
\[ y \approx 15 \cdot 1.43 - 19.6 \cdot (1.43)^2 \approx 21.45 - 40.00 \approx -18.55 \, \text{m} \]
- Since the height is measured from the hill, the effective height is \( 10 - 18.55 \approx 5.55 \, \text{m} \).

### Final Results
1. **Time interval between firings**: \( \Delta t \approx 1.43 \, \text{s} \)
2. **Coordinates of point P**: \( (12.35 \, \text{m}, 5.55 \, \text{m}) \)