Two guns, situated on the top of a hill of height $10\,\text{m}$, fire one shot each with the same speed $5\sqrt{3}\,\text{m/s}$ at some interval of time. One gun fires horizontally and other fires upwards at an angle of $60^\circ$ with the horizontal. The shots collide in air at a point $P$. Find (i) the time-interval between the firings, and (ii) the coordinates of the point $P$. Take origin of the coordinate system at the foot of the hill right below the muzzle and trajectories in $x$-$y$ plane.

Step 1: Define the coordinate system and initial conditions.

The origin $(0,0)$ is at the foot of the hill, which places the muzzle at $(0, 10)\text{ m}$.

$$g = 10\text{ m/s}^2$$ $$u = 5\sqrt{3}\text{ m/s}$$

Step 2: Write the equations of motion for Gun A.

Gun A fires horizontally, and we let $t$ be its time of flight up to the collision point.

$$x_A = 5\sqrt{3}t$$ $$y_A = 10 - 5t^2$$

Step 3: Write the equations of motion for Gun B.

Gun B fires at $60^\circ$ above the horizontal, and we let $\tau$ be its time of flight up to the collision point.

$$x_B = \left(5\sqrt{3}\cos 60^\circ\right)\tau = \frac{5\sqrt{3}}{2}\tau$$ $$y_B = 10 + \left(5\sqrt{3}\sin 60^\circ\right)\tau - \frac{1}{2}g\tau^2$$ $$y_B = 10 + \frac{15}{2}\tau - 5\tau^2$$

Step 4: Equate the horizontal coordinates at the collision point.

Since the shots collide at point $P$, their horizontal positions must be equal.

$$x_A = x_B$$ $$5\sqrt{3}t = \frac{5\sqrt{3}}{2}\tau$$ $$\tau = 2t$$

Step 5: Equate the vertical coordinates at the collision point.

The vertical positions must also be equal at the point of collision.

$$y_A = y_B$$ $$10 - 5t^2 = 10 + \frac{15}{2}\tau - 5\tau^2$$

Substitute $\tau = 2t$ into the vertical equation.

$$10 - 5t^2 = 10 + \frac{15}{2}(2t) - 5(2t)^2$$ $$10 - 5t^2 = 10 + 15t - 20t^2$$ $$15t^2 - 15t = 0$$

Since $t > 0$ for a collision after firing, we solve for $t$.

$$t = 1\text{ s}$$ $$\tau = 2\text{ s}$$

Step 6: Calculate the time-interval between the firings.

The time-interval $\Delta t$ is the difference between the flight times of the two shots.

$$\Delta t = \tau - t$$ $$\Delta t = 2 - 1 = 1\text{ s}$$

Step 7: Determine the coordinates of the collision point $P$.

Substitute $t = 1\text{ s}$ into the position equations of Gun A.

$$x_P = 5\sqrt{3}(1) = 5\sqrt{3}\text{ m}$$ $$y_P = 10 - 5(1)^2 = 5\text{ m}$$

Answer: (i) The time-interval between the firings is $1\text{ s}$. (ii) The coordinates of the point $P$ are $(5\sqrt{3}\text{ m}, 5\text{ m})$.