Two guns situated on the top of a hill of height 10 m fire one shot each with the same speed 5√3 m/s

Q: Wait, why do we assume the second bullet was fired first?

Look at their horizontal velocities. The first bullet travels horizontally at $5\sqrt{3}\text{ m/s}$. The second bullet's horizontal speed is $5\sqrt{3} \cos 60^\circ$, which is half as fast. To cover the exact same horizontal distance $x$ to reach the collision point, the slower bullet must have been traveling for a longer time. Hence, $t_2 > t_1$, meaning it was fired earlier.

Q: Why is the initial height $10\text{ m}$ included in the $y$ equation?

The problem states the origin $(0,0)$ is at the *foot* of the hill. Since the guns are on top of the $10\text{ m}$ hill, their initial $y$-coordinate is $+10\text{ m}$. If you placed the origin at the top of the hill, the initial height would be $0$, but you would have to adjust the final answer to match the coordinate system requested by the question.

Q: Can we solve this using relative velocity?

Yes, absolutely! If you observe the second bullet from the frame of the first bullet, the relative acceleration is zero (since both experience $g$ downwards). The relative velocity must be directed exactly towards the initial position of the first bullet for a collision to occur. However, because they are fired at *different* times, the relative velocity approach becomes slightly more complex to set up than simply equating their coordinates.

Q: What happens to the $10$ in the vertical equation?

Since both bullets are fired from the exact same height of $10\text{ m}$, the term $10$ appears on both sides of the equation $y_1 = y_2$. When you subtract $10$ from both sides, it perfectly cancels out. This physically means the collision condition depends only on their relative vertical displacements from the firing point.

Q: Why did we reject $t_1 = 0$ when solving the quadratic equation?

Mathematically, $15t_1^2 = 15t_1$ has two roots: $t_1 = 0$ and $t_1 = 1$. Physically, $t_1 = 0$ represents the exact moment the first bullet is fired. At that instant, the second bullet is already in the air (since $t_2 = 2\text{ s}$), so they are not at the same location. The collision happens later, at $t_1 = 1\text{ s}$.

Question

Two guns situated on the top of a hill of height $10	ext{ m}$ fire one shot each with the same speed $5\sqrt{3}	ext{ m/s}$ 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 point $P$ ($g = 10	ext{ m/s}^2$). Find
(a) the time interval between the firings and
(b) 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.

Sigma Percentile · Accepted Answer

### The Thrill of the Mid-Air Collision

Imagine standing on the edge of a $10	ext{ m}$ high cliff, holding two projectile launchers. You fire the first one perfectly horizontally. Then, after a brief, calculated pause, you fire the second one upwards at a steep $60^\circ$ angle. Both projectiles leave the barrel at the exact same speed of $5\sqrt{3}	ext{ m/s}$.

Your mission? To determine the exact time delay required between the shots so that they obliterate each other in a spectacular mid-air collision, and to find the exact coordinates of that impact. This is a classic kinematics puzzle that tests your ability to decouple horizontal and vertical motion.

### Analyzing the Setup

Let's establish our coordinate system. We place the origin $(0,0)$ at the foot of the hill. This means both projectiles are launched from the coordinates $(0, 10)$.

For a collision to occur, both projectiles must arrive at the exact same point $P(x, y)$ at the exact same moment in time. Let's define $t_1$ as the time the first projectile spends in the air, and $t_2$ as the time the second projectile spends in the air.

Because the second projectile is fired at an angle, its horizontal velocity component is smaller than the first projectile's. To cover the same horizontal distance $x$, the second projectile must be in the air for a longer duration. This tells us intuitively that $t_2 > t_1$, meaning the second projectile must be fired *first*.

### The Master Equations

We begin by analyzing the horizontal motion. Since there is no air resistance, the horizontal velocity remains constant. We can equate the horizontal displacements of both projectiles:

$$u_1 t_1 = (u_2 \cos 60^\circ) t_2$$

Substituting the given initial velocities ($u_1 = u_2 = 5\sqrt{3}	ext{ m/s}$) and the value of $\cos 60^\circ = \frac{1}{2}$, we get:

$$5\sqrt{3} t_1 = 5\sqrt{3} \left(\frac{1}{2}ight) t_2$$

The $5\sqrt{3}$ beautifully cancels out from both sides, leaving us with a powerful and simple relationship between the flight times:

$$t_2 = 2t_1$$

Next, we analyze the vertical motion. Both projectiles start at a height of $10	ext{ m}$ and must end up at the same final height $y$. We write the vertical position equations for both:

$$y = 10 - \frac{1}{2} g t_1^2$$

$$y = 10 + (u_2 \sin 60^\circ) t_2 - \frac{1}{2} g t_2^2$$

Equating these two expressions gives us our master vertical equation:

$$10 - \frac{1}{2} g t_1^2 = 10 + (u_2 \sin 60^\circ) t_2 - \frac{1}{2} g t_2^2$$

### The Time Interval

Notice how the initial height of $10	ext{ m}$ cancels out immediately. This makes physical sense; the collision condition depends only on their relative vertical displacements from the launch point.

Let's substitute $g = 10	ext{ m/s}^2$, $u_2 = 5\sqrt{3}	ext{ m/s}$, and $\sin 60^\circ = \frac{\sqrt{3}}{2}$ into the equation:

$$-5 t_1^2 = \left(5\sqrt{3} \cdot \frac{\sqrt{3}}{2}ight) t_2 - 5 t_2^2$$

$$-5 t_1^2 = 7.5 t_2 - 5 t_2^2$$

Now, we bring in our golden relationship, $t_2 = 2t_1$, and substitute it into this equation to eliminate $t_2$:

$$-5 t_1^2 = 7.5(2t_1) - 5(2t_1)^2$$

$$-5 t_1^2 = 15 t_1 - 20 t_1^2$$

Rearranging the terms to group the quadratics yields:

$$15 t_1^2 = 15 t_1$$

Since the time of collision $t_1$ cannot be zero (that would be the moment of launch), we can safely divide both sides by $15 t_1$, giving us:

**$t_1 = 1	ext{ s}$**

Using our relationship $t_2 = 2t_1$, we find that $t_2 = 2	ext{ s}$. The time interval between the firings is simply the difference between their flight times:

**$\Delta t = t_2 - t_1 = 1	ext{ s}$**

### The Final Coordinates

With the flight time $t_1$ known, finding the exact coordinates of the collision point $P(x, y)$ is a breeze. We simply plug $t_1 = 1	ext{ s}$ back into our original position equations for the first projectile.

For the horizontal coordinate:

$$x = u_1 t_1 = 5\sqrt{3}(1) = 5\sqrt{3}	ext{ m}$$

For the vertical coordinate:

$$y = 10 - \frac{1}{2} g t_1^2 = 10 - \frac{1}{2}(10)(1)^2 = 10 - 5 = 5	ext{ m}$$

Thus, the spectacular mid-air collision occurs exactly at the coordinates **$(5\sqrt{3}	ext{ m}, 5	ext{ m})$**. A perfect bullseye achieved through the elegant application of kinematics!

Visualized Solution

Visualizing the Setup

The Condition for Collision

Equating Horizontal Displacements

Relationship Between $t_{1}$ and $t_{2}$

Equating Vertical Displacements

Simplifying the Vertical Equation

Solving for $t_{1}$

Calculating $t_{1}$ and $t_{2}$

Coordinates of the Collision Point $P$

The Sigma Insight: Projectile Motion

The Thrill of the Mid-Air Collision

Analyzing the Setup

The Master Equations

The Time Interval

The Final Coordinates

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