Projectile Motion

Cheatsheet Content

### Introduction to Projectile Motion - **Definition:** The motion of an object thrown or projected into the air, subject only to the acceleration of gravity. - **Assumptions:** - Air resistance is negligible. - The acceleration due to gravity ($g$) is constant and acts downwards ($\approx 9.81 \, \text{m/s}^2$). - The Earth's curvature is negligible. - The object's rotation is negligible. - **Key Principle:** Horizontal and vertical motions are independent. ### Kinematic Equations (Constant Acceleration) - These equations are fundamental for solving projectile motion problems. - **General Forms:** 1. $$v = v_0 + at$$ 2. $$x = x_0 + v_0t + \frac{1}{2}at^2$$ 3. $$v^2 = v_0^2 + 2a(x - x_0)$$ 4. $$x - x_0 = \frac{1}{2}(v_0 + v)t$$ - **Variables:** - $v$: final velocity - $v_0$: initial velocity - $a$: acceleration - $x$: final position - $x_0$: initial position - $t$: time ### Horizontal Motion (x-direction) - **Acceleration:** $a_x = 0$ (constant velocity) - **Equations:** 1. $$v_x = v_{0x}$$ 2. $$x = x_0 + v_{0x}t$$ - Here, $v_{0x}$ is the initial horizontal velocity component. - **Note:** The horizontal velocity remains constant throughout the flight, assuming no air resistance. ### Vertical Motion (y-direction) - **Acceleration:** $a_y = -g$ (downwards, where $g \approx 9.81 \, \text{m/s}^2$) - **Equations:** 1. $$v_y = v_{0y} - gt$$ 2. $$y = y_0 + v_{0y}t - \frac{1}{2}gt^2$$ 3. $$v_y^2 = v_{0y}^2 - 2g(y - y_0)$$ - Here, $v_{0y}$ is the initial vertical velocity component. - **At Maximum Height:** $v_y = 0$. ### Initial Velocity Components - If an object is launched with initial speed $v_0$ at an angle $\theta$ above the horizontal: - **Horizontal Component:** $v_{0x} = v_0 \cos\theta$ - **Vertical Component:** $v_{0y} = v_0 \sin\theta$ - **If launched horizontally:** $\theta = 0^\circ$, so $v_{0x} = v_0$ and $v_{0y} = 0$. ### Key Projectile Motion Parameters #### Time of Flight ($T$) - The total time the projectile spends in the air. - **For launch from ground to ground ($y_0 = y = 0$):** $$T = \frac{2v_{0y}}{g} = \frac{2v_0 \sin\theta}{g}$$ - **General case:** Use $y = y_0 + v_{0y}t - \frac{1}{2}gt^2$ and solve for $t$ when $y$ is known. #### Maximum Height ($H$) - The highest vertical position reached by the projectile. - Occurs when $v_y = 0$. - **For launch from ground to ground ($y_0 = 0$):** $$H = \frac{v_{0y}^2}{2g} = \frac{(v_0 \sin\theta)^2}{2g}$$ - **General case:** Use $v_y^2 = v_{0y}^2 - 2g(y - y_0)$ with $v_y = 0$. #### Horizontal Range ($R$) - The total horizontal distance covered by the projectile. - **For launch from ground to ground ($y_0 = y = 0$):** $$R = v_{0x}T = v_0 \cos\theta \left( \frac{2v_0 \sin\theta}{g} \right) = \frac{v_0^2 \sin(2\theta)}{g}$$ - **Maximum Range:** Occurs at $\theta = 45^\circ$ for a given $v_0$ (when $y_0 = y = 0$). #### Velocity at any time ($t$) - **Horizontal Velocity:** $v_x(t) = v_{0x}$ - **Vertical Velocity:** $v_y(t) = v_{0y} - gt$ - **Magnitude of Velocity:** $|v(t)| = \sqrt{v_x(t)^2 + v_y(t)^2}$ - **Direction of Velocity:** $\phi = \arctan\left(\frac{v_y(t)}{v_x(t)}\right)$ (angle with horizontal) ### Equation of Trajectory - Describes the path of the projectile in the x-y plane. - By eliminating $t$ from the position equations: $$y = x \tan\theta - \frac{gx^2}{2(v_0 \cos\theta)^2}$$ - This is the equation of a parabola. ### Special Cases #### Horizontal Launch from a Height ($v_{0y} = 0$) - **Initial conditions:** $v_0$ horizontal, $y_0 = H_{initial}$, $x_0 = 0$. - **Time to hit ground:** $t = \sqrt{\frac{2H_{initial}}{g}}$ (from $y = H_{initial} - \frac{1}{2}gt^2$ with $y=0$) - **Horizontal distance (range):** $R = v_0 t = v_0 \sqrt{\frac{2H_{initial}}{g}}$ - **Final vertical velocity:** $v_y = -gt = - \sqrt{2gH_{initial}}$ - **Final speed:** $|v| = \sqrt{v_0^2 + v_y^2}$ #### Projectile Launched Upwards from a Height - Follows general equations, but $y_0 \ne 0$. - Often requires using the quadratic formula to solve for time $t$ when $y$ is given. $$- \frac{1}{2}gt^2 + v_{0y}t + (y_0 - y) = 0$$ ### Problem Solving Strategy 1. **Draw a Diagram:** Include initial position, velocity, angle, and coordinate system. 2. **Define Coordinate System:** Choose origin ($x_0, y_0$) and positive directions (e.g., up = positive y, right = positive x). 3. **List Knowns & Unknowns:** Separate into horizontal (x) and vertical (y) components. - $v_{0x} = v_0 \cos\theta$ - $v_{0y} = v_0 \sin\theta$ - $a_x = 0$ - $a_y = -g$ (or $+g$ if positive y is downwards) 4. **Choose Appropriate Equations:** Select kinematic equations that relate the knowns to the unknowns. 5. **Solve for Time First:** Time is the common link between horizontal and vertical motion. Often, solving for time in one direction (e.g., vertical to find time to max height or time to hit ground) allows you to solve for unknowns in the other direction. 6. **Substitute and Calculate:** Plug in values and ensure units are consistent.