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 point `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 step by step, we will analyze the motion of both shots fired from the guns and find the time interval between their firings and the coordinates of the collision point \( P \).

### Step 1: Understand the motion of the shots
- The first gun fires horizontally with a speed of \( 5\sqrt{3} \, \text{m/s} \).
- The second gun fires at an angle of \( 60^\circ \) with the horizontal, also with a speed of \( 5\sqrt{3} \, \text{m/s} \).

### Step 2: Resolve the velocity of the second shot
- The horizontal component of the velocity of the second shot:
\[ u_{x2} = 5\sqrt{3} \cos(60^\circ) = 5\sqrt{3} \cdot \frac{1}{2} = \frac{5\sqrt{3}}{2} \, \text{m/s} \]
- The vertical component of the velocity of the second shot:
\[ u_{y2} = 5\sqrt{3} \sin(60^\circ) = 5\sqrt{3} \cdot \frac{\sqrt{3}}{2} = \frac{15}{2} \, \text{m/s} \]

### Step 3: Define the time of firing
- Let the first shot be fired at \( t = 0 \).
- Let the second shot be fired at \( t = t_0 \).
- The time of flight for the first shot when it collides is \( t \).
- The time of flight for the second shot when it collides is \( t - t_0 \).

### Step 4: Write the equations of motion
- For the first shot (horizontal):
\[ x_1 = 5\sqrt{3} t \]
\[ y_1 = 10 - \frac{1}{2} g t^2 = 10 - \frac{5}{2} t^2 \]

- For the second shot (at \( 60^\circ \)):
\[ x_2 = \frac{5\sqrt{3}}{2} (t - t_0) \]
\[ y_2 = 10 + \frac{15}{2} (t - t_0) - \frac{1}{2} g (t - t_0)^2 \]

### Step 5: Set the x-coordinates equal for collision
Since both shots collide at point \( P \):
\[ 5\sqrt{3} t = \frac{5\sqrt{3}}{2} (t - t_0) \]
Solving this equation:
\[ 10t = 5(t - t_0) \]
\[ 10t = 5t - 5t_0 \]
\[ 5t = -5t_0 \implies t = 2t_0 \]

### Step 6: Set the y-coordinates equal for collision
Now equate the y-coordinates:
\[ 10 - \frac{5}{2} t^2 = 10 + \frac{15}{2} (t - t_0) - \frac{1}{2} g (t - t_0)^2 \]
Substituting \( g = 10 \, \text{m/s}^2 \):
\[ -\frac{5}{2} t^2 = \frac{15}{2} (t - t_0) - 5(t - t_0)^2 \]
Substituting \( t = 2t_0 \):
\[ -\frac{5}{2} (2t_0)^2 = \frac{15}{2} (2t_0 - t_0) - 5(2t_0 - t_0)^2 \]
\[ -10t_0^2 = \frac{15}{2} t_0 - 5t_0^2 \]
\[ -10t_0^2 + 5t_0^2 = \frac{15}{2} t_0 \]
\[ -5t_0^2 = \frac{15}{2} t_0 \]
\[ -10t_0 = 15 \implies t_0 = 1 \, \text{s} \]

### Step 7: Find the coordinates of point \( P \)
Now substituting \( t = 2t_0 = 2 \, \text{s} \) into the equations for \( P \):
- For \( x \):
\[ x = 5\sqrt{3} \cdot 2 = 10\sqrt{3} \, \text{m} \]
- For \( y \):
\[ y = 10 - \frac{5}{2} (2)^2 = 10 - 10 = 0 \, \text{m} \]

### Final Result
The coordinates of point \( P \) are \( (10\sqrt{3}, 0) \).