Comprehensive Study Notes on Horizontally Launched Projectile Motion

Fundamentals of Horizontally Launched Projectiles and Free Fall

Projectile motion is defined as motion along a curved path where gravity is the only force acting upon the object. The standard acceleration due to gravity on Earth is approximately $9.81\,m/s^2$, often rounded to $10\,m/s^2$ for conceptual simplicity. When air resistance is neglected, the horizontal velocity ($V_x$) of a projectile remains constant throughout its flight. In contrast, the vertical velocity ($V_y$) increases by approximately $10\,m/s$ every second due to the force of gravity. A critical principle of kinematics is that the horizontal and vertical components of motion are entirely independent of one another. This independence is demonstrated by comparing an object dropped straight down (free fall) with an object launched horizontally from the same height. Because gravity is the only force pulling both objects downward, they accelerate at the same rate and strike the surface at the identical time.

Kinematic Analysis of a Projectile Launched at $20\,m/s$

For a projectile launched horizontally at an initial velocity ($V_x$) of $20\,m/s$, we can determine the resultant speed and angle at specific intervals. At $1\,s$, the horizontal velocity is $20\,m/s$ and the vertical velocity is $10\,m/s$. The resultant speed is calculated using the Pythagorean theorem: $V = \sqrt{20^2 + 10^2} = 22.4\,m/s$. The angle relative to the x-axis is determined by $\theta = \tan^{-1}(\frac{10}{20}) = 27^{\circ}$ below the positive x-axis, which can also be expressed as $333^{\circ}$ from the x-axis. At $2\,s$, with $V_x = 20\,m/s$ and $V_y = 20\,m/s$, the resultant speed is $V = \sqrt{20^2 + 20^2} = 28.3\,m/s$ at an angle of $45^{\circ}$ below the positive x-axis (or $315^{\circ}$).

As the projectile continues its descent, at $3\,s$, the vertical velocity reaches $30\,m/s$. The resultant speed is $V = \sqrt{20^2 + 30^2} = 36.1\,m/s$. The angle is calculated as $\theta = \tan^{-1}(\frac{30}{20}) = 56^{\circ}$ below the positive x-axis (or $304^{\circ}$). Finally, at $4\,s$, the vertical velocity is $40\,m/s$, resulting in a speed of $V = \sqrt{20^2 + 40^2} = 44.7\,m/s$. The angle at this point is $\theta = \tan^{-1}(\frac{40}{20}) = 63^{\circ}$ below the positive x-axis (or $297^{\circ}$).

Mathematical Framework for Projectile Motion Components

To analyze projectile motion problems effectively, horizontal and vertical components must be considered separately. In the x-direction, $\Delta x$ represents the horizontal displacement or range, and $V_x$ is the constant horizontal velocity. Notably, horizontal acceleration ($a_x$) is always $0$ because $V_x$ does not change. In the y-direction, $\Delta y$ represents the vertical displacement or height, and $V_y$ is the vertical velocity. The vertical acceleration ($a_y$) on Earth is always the acceleration due to gravity, defined as $-9.8\,m/s^2$. When an object is launched horizontally, the initial vertical velocity ($V_{oy}$) is equal to $0$.

Example A: Rock Launched Horizontally from a Building

Consider a rock of mass $m$ thrown horizontally off a building of height $h = 15\,m$ with an initial speed $V_i = 10\,m/s$. To find the time ($t$) it takes for the object to hit the ground, we utilize the vertical displacement formula $\Delta y = V_{oy}t + \frac{1}{2}a t^2$. Since the rock is thrown horizontally, $V_{oy} = 0$, simplifying the equation to $\Delta y = \frac{1}{2}a t^2$. Solving for $t$ gives $t = \sqrt{\frac{2\Delta y}{a}}$. Using the values $\Delta y = -15\,m$ and $a = -9.8\,m/s^2$, the time is calculated as $t = \sqrt{\frac{2(-15)}{-9.8}} = 1.75\,s$. It is important to note that the time in a projectile path is equal for both the vertical and horizontal portions of the motion.

The horizontal distance traveled, or the range ($\Delta x$), is found using the formula $\Delta x = V_{ox}t + \frac{1}{2}a_x t^2$. Given that $a_x = 0$, this simplifies to $\Delta x = V_{ox}t$. Substituting the known values, $\Delta x = (10\,m/s)(1.75\,s) = 17.5\,m$. To determine the final velocity with which the rock hits the ground, we first find the final vertical velocity: $V_f = V_o + at = (0) + (-9.8)(1.75\,s) = -17.15\,m/s$. The magnitude of the resultant final velocity is $V_f = \sqrt{V_y^2 + V_x^2} = \sqrt{(-17.15)^2 + 10^2} = 19.9\,m/s$. The direction is $\theta = \tan^{-1}(\frac{17.15}{10}) = 60^{\circ}$ below the positive x-axis.

Conceptual Variations in Height and Speed

If the height of the cliff in the previous example were doubled, the rock would travel a greater horizontal distance. This occurs because a larger vertical displacement ($\Delta y$) results in a longer time to fall, and during that longer fall time, the constant horizontal velocity has more time to cover a greater distance. Conversely, if the initial horizontal speed of the object were doubled, the time the rock stays in the air would remain unchanged. This is because time in the air is strictly based on vertical motion and height; it is not affected by horizontal speed.

Example B: Crate Dropped from a Moving Plane

A plane flying horizontally at a constant speed of $120\,m/s$ at a height of $300\,m$ drops a crate. The initial conditions are $V_{ox} = 120\,m/s$, $V_{oy} = 0$, $a_x = 0$, $a_y = -9.8\,m/s^2$, and $\Delta y = -300\,m$. As the crate falls, its horizontal speed remains $120\,m/s$ because there is no horizontal acceleration. To find how long it takes for the crate to hit the ground, we use $t = \sqrt{\frac{2\Delta y}{a_y}} = \sqrt{\frac{2(-300)}{-9.8}} = 7.8\,s$. The horizontal distance from the release point where the crate hits the ground is calculated as $\Delta x = V_{ox}t = (120\,m/s)(7.8\,s) = 936\,m$.

To find the crate's velocity immediately before impact, we first calculate the final vertical velocity: $V_{yf} = V_o + at = 0 + (-9.8)(7.8) = -76.44\,m/s$. We then determine the resultant velocity magnitude: $V_f = \sqrt{V_y^2 + V_x^2} = \sqrt{76.44^2 + 120^2} = 142\,m/s$. The direction of the final velocity is found by $\theta = \tan^{-1}(\frac{76.44}{120}) = 32.5^{\circ}$ below the positive x-axis.