To solve this problem, we need to analyze the motion of both projectiles fired from the guns. One gun fires horizontally, while the other fires at an angle of 60 degrees. We will break down the problem into two parts: finding the time interval between the firings and determining the coordinates of the collision point P.

1. Time Interval Between the Firings

Let’s denote the speed of the shots as v m/s. The first gun fires horizontally, and the second gun fires at an angle of 60 degrees. We will use the equations of motion to find the time it takes for each projectile to reach the same height.

Projectile Motion of the Horizontal Shot

The horizontal gun fires its projectile with an initial velocity v in the horizontal direction. The vertical motion is influenced by gravity. The time t_1 it takes for the projectile to fall from a height of 10 m can be calculated using the equation:

• y = h - \frac{1}{2}gt^2

Here, y is the vertical position, h is the initial height (10 m), g is the acceleration due to gravity (approximately 9.81 m/s²), and t is the time in seconds. Setting y to 0 (the ground level), we have:

• 0 = 10 - \frac{1}{2}(9.81)t_1^2

Rearranging gives:

• \frac{1}{2}(9.81)t_1^2 = 10
• t_1^2 = \frac{20}{9.81}
• t_1 = \sqrt{\frac{20}{9.81}} \approx 1.43 \text{ seconds}

Projectile Motion of the Angled Shot

For the second gun, which fires at an angle of 60 degrees, we can break the initial velocity into horizontal and vertical components:

• v_x = v \cdot \cos(60°) = \frac{v}{2}
• v_y = v \cdot \sin(60°) = \frac{v\sqrt{3}}{2}

The vertical motion of this projectile can be described by the equation:

• y = v_y t - \frac{1}{2}gt^2

Setting y to 0 at the point of collision, we have:

• 0 = \frac{v\sqrt{3}}{2}t_2 - \frac{1}{2}(9.81)t_2^2

Factoring out t_2 gives:

• t_2 \left(\frac{v\sqrt{3}}{2} - \frac{1}{2}(9.81)t_2\right) = 0

This yields two solutions: t_2 = 0 (the moment of firing) and:

• t_2 = \frac{v\sqrt{3}}{9.81}

Finding the Time Interval

To find the time interval Δt between the firings, we set the two times equal to each other at the point of collision:

• t_1 + Δt = t_2

Substituting the expressions we found:

• 1.43 + Δt = \frac{v\sqrt{3}}{9.81}

Solving for Δt gives:

• Δt = \frac{v\sqrt{3}}{9.81} - 1.43

2. Coordinates of the Collision Point P

Now, we need to find the coordinates of point P where the two projectiles collide. We will calculate the horizontal distance traveled by both projectiles at the time of collision.

Horizontal Distance of the Horizontal Shot

The horizontal distance x_1 traveled by the first gun is:

• x_1 = v \cdot t_1

Horizontal Distance of the Angled Shot

The horizontal distance x_2 traveled by the second gun is:

• x_2 = v_x \cdot t_2 = \frac{v}{2} \cdot \frac{v\sqrt{3}}{9.81}

Setting the Distances Equal

Since both projectiles collide at the same point, we can set x_1 equal to x_2:

• v \cdot t_1 = \frac{v}{2} \cdot \frac{v\sqrt{3}}{9.81}

After canceling v from both sides (assuming v ≠ 0), we can solve for the coordinates of point P:

• x = v \cdot t_1
• y = 0 (since it’s at ground level)

In summary, the time interval between the firings is given by:

• Δt = \frac{v\sqrt{3}}{9.81} - 1.43

And the coordinates of point P, where the shots collide, can be expressed as:

• (x, 0), where x = v \cdot 1.43

By substituting the value of v, you can