91Ó°ÊÓ

When dealing with horizontal motion in physics, it's important to understand that this type of motion occurs when an object moves along a flat plane, parallel to the ground. In the problem of the bullet being fired from the rifle, the bullet travels in a straight line horizontally towards the target. This motion can be described by only one component of velocity: the horizontal velocity, denoted as \(v_{i_x}\).
This velocity remains constant throughout the bullet's journey toward the bull's-eye because no horizontal forces (friction or air resistance are typically neglected in simple physics problems) are acting on it. In our problem, this constant horizontal velocity is \(670 \, \text{m/s}\).
We used this value together with the time of flight to find the horizontal distance using the equation:

• \( x = v_{i_x} \cdot t \)

This principle of constant horizontal velocity and the simple multiplication with the time of travel provides the means to calculate how far the bullet travels horizontally.

Vertical motion in physics deals with how objects move up or down under the influence of forces like gravity. In our exercise, vertical motion is crucial because gravity causes the bullet to drop as it travels horizontal distance.
The problem tells us that the bullet strikes 0.025 meters below the bull's-eye. Here, the bullet's initial vertical velocity is 0 m/s, as the rifle is fired horizontally. The force causing the bullet to drop is gravity, which accelerates objects at a rate of \(9.81 \, \text{m/s}^2\). To find the time \(t\) it takes for the bullet to travel the vertical distance, we use the vertical displacement formula:

• \( y = v_{i_y}t + \frac{1}{2}gt^2 \)

Given \(v_{i_y} = 0\), this simplifies to solving for \(t\) as:

• \( 0.025 = \frac{1}{2}(9.81)t^2 \)

We found that \(t \approx 0.071 \, \text{s}\). This time is then used to understand both the vertical and horizontal motions of the bullet.

Projectile motion is the combination of both horizontal and vertical motions. This type of motion occurs when an object is launched into the air and moves along a curved path under the influence of gravity. The motion path is typically a parabola.
In the problem, the bullet follows a projectile path, even though it's initially fired straight horizontally. Hence, understanding the separate horizontal and vertical components is crucial. Gravity affects only the vertical motion, causing the bullet to drop over distance, but doesn't influence horizontal speed, keeping it at 670 m/s.
By combining the calculated time from the vertical motion with the initial horizontal velocity, you can determine the total horizontal distance covered. For projectiles, knowing how these components work together helps in predicting where the projectile will land, as it did in calculating the \(47.57 \, \text{meters}\) for the bullet to strike below the bull's-eye.