--- ## Projectile motion - A projectile is an object that **is traveling with a constant horizontal/x velocity.** The y component of a projectiles motion **always has negative acceleration** due to gravity. For basic projectile motion we assume that the projectile is in a vacuum and therefore does not feel the effects of wind resistance. - The ideal angle for a projectile to be shot so that it travels furthest is **45 degrees.** - When the projectile is at the ==**vertex of its parabolic arc, its y component of velocity is equal to zero since at that moment the projectile is neither moving up nor down. However, even at this vertex the projectiles y component still experiences negative acceleration due to gravity.**== - At the very moment a projectile is shot, both the x and y components of it's velocity are positive. - If there are no forces influencing the projectile when it is in flight, the y component of its acceleration is **always -9.8m/s.** - Assuming the projectile is shot upwards in a parabolic arc on a perfectly flat plane the overall **speed/V magnitude** of the projectile *(including both its x and y components)* is same at both the moment its shot and at the moment it lands. The overall speed of the projectile is the least when it is at its vertex since at that moment only, the **y component of its velocity does not contribute to its overall speed.** - The **y velocity of a projectile depends on whether it is moving up or down**, of course eventually it will always start moving down and hit the ground due to gravity. - The x component of velocity for a projectile and the y component of its velocity can be calculated using the following equations. Given that you know the projectiles initial velocity and angle. $\large v_x=v\cdot cos(\theta)\space\space\space v_y=v\cdot sin(\theta)$ - By breaking the velocity/motion of the projectile into its individual components we can calculate its position at different points in time using the basic kinematic equations described on the [[Velocity]] page. - We use the **velocity + acceleration kinematic equation to calculate how a projectile's y position changes over time**. We have to do this because the y component of a projectile's velocity experiences **acceleration due to gravity**. $\Huge\boxed{\normalsize\text{Y coordinate of a projectile: }\; y_{\small f}=y_{\small i}+t\cdot v_{\small i\,y}-\frac{1}{2}gt^2}$ $\newline$ - Since the x component **of a particle's velocity does not have any acceleration we can find the x coordinate of a particle's position using the simple velocity vs time equation:** $\Huge\boxed{\normalsize\text{X coordinate of a projectile: }\;\;x_{\small f}=x_{\small i}+t\cdot v_{i\,x}}$ $\newline$ - Based on these and other kinematic equations *(see [[Velocity]])* we can derive a few simple formulas for the range, max height, flight time and time to max height of a projectile. These equations **are extremely useful to use in conjunction with the above kinematic equations since they provide you information about where a projectile is at a specific time throughout its flight.** $\newline$ $\text{Range: }R=v_x\cdot t_f=\left({\color{LightCyan}\frac{v_i^2\cdot sin(2\theta)}{g}}\right)\space\space\space\space\space\text{Flight time: }t_f=\frac{2v_{i,y}}{g}=\left({\color{LightCyan}\frac{2v_i\cdot sin(\theta)}{g}}\right)$ $\text{Max height: }H=\left({\color{LightCyan}\frac{v_i^2\cdot sin^2(\theta)}{2\cdot g}}\right)\space\space\space\space\space\space\space\space\text{Time to max height: }\Delta t=\color{LightCyan}\sqrt{\frac{2\Delta y}{g}}$ $\small\color{grey}\Delta y\text{ is the max height of projectile}$ ##### *Example problem* >[!example] >- A ball is thrown horizontally from a 20 m high building with a speed of 9.0 meters per second. How far from the base of the building does the ball hit the ground? > >$\text{\small At what time does the ball hit the ground? }\rightarrow 0=20-\frac{1}{2}9.8\cdot t^2\longrightarrow\sqrt{\frac{20}{0.5\cdot 9.8}}=t=2.02s$ >$\newline$ >$\text{\small How far from the base does it travel? }\rightarrow\Delta x=9.0\cdot 2.02=\color{Bisque}18.18m$ --- ## Common questions: **==What is the overall speed/velocity of the projectile?==** - If you know the angle the projectile was shot at and the velocity of one of its components you can use either $\Large\frac{v_y}{sin(\theta)}$ or $\Large\frac{v_x}{cos(\theta)}$ to find the distance of the hypotenuse which is equal to the overall velocity. - Imagine you are trying to find the velocity of a bullet and you only know the distance to the target and how far the bullet dropped. The first step is to find the **time it took the bullet to reach the target**, to do this re-arrange the acceleration/position equation: $x_f=x_i+\frac{1}{2}at^2$ to solve **for t** *(set $x_f$ equal to the bullet drop and $x_i$ to 0)*. Now, re-arrange the velocity/position equation to solve for velocity, setting $x_i$ to 0 and $x_f$ to the distance of the target. - By re-arranging the velocity/position equation to solve for velocity you are essentially asking the question: *How fast does the bullet need to be for it to travel x distance over t time.* - If you are **trying to find the speed of the projectile at some specific time** find the velocity of its **x and y components, then use the Pythagorean theorem to combine them** giving you the *overall* velocity at that time. $v=\sqrt{v_x^2+v_y^2}$ - If you know the **distance** the projectile was shot and the angle, you can re-arrange the range formula to solve for v. The re-arranged equation looks like this: $v=\sqrt{\frac{\Delta x\cdot g}{sin(2\theta)}}$ $\newline$ **==What is the position of the object?==** - If you are given an angle and a velocity, use the angle along with $v\cdot sin(\theta)$ and $v\cdot cos(\theta)$ to find the x and y components of the object's velocity. Then using the time of flight equation figure out how long $t$ the object was in the air for. Using this information and the standard $x_f=x_i+vt$ you can find the position *(either x or y)* of the ball at any point in time. - If you are finding the y position of a projectile that has been shot up into the air with some velocity $v$ you must remember to always account for the acceleration due to gravity. Use the equation $x=vt+\frac{1}{2}at^2$ to find the position when both velocity and acceleration are involved. **==How long is the object in the air for?==** - This question is easy to solve if you are shooting the projectile over a **flat plane** since you can use the air time equation outlined earlier in this section. However, If you are shooting a projectile off the **top of a building you cannot use that equation!**. Instead you use the acceleration/position equation, re-arranging it to solve for t and setting $x_i$ equal to the height of the building. Assuming you are throwing the projectile off the building horizontally. - If you are shooting the projectile off a building at some angle up into the air you will have to solve first **for the amount of time the projectile takes to reach its apex**. This can be done using the standard air time equation since the projectile never goes *below the person/thing that shot it*. Make sure to **divide the solution $t$ in half**. After this you can find the time it would take a projectile that you threw horizontally off a building that's at the height of the original projectile's vertex to reach the ground *(use the method outlined above)*. By adding these two times together you have your result! - The equation for the flight time of a projectile launched above the ground at an angle is shown in the section below. --- ## More useful equations: >[!quote] $\text{Flight time when with height }h\text{:}\;\;\;\Huge\boxed{\normalsize\color{Lightcyan}t=\frac{v_i\cdot sin(\theta)+\sqrt{\Big(v_i\cdot sin(\theta)\Big)^2+2gh}}{g}}$ $\text{Initial velocity from height }h\text{ angle, and range }R\text{:}\;\;\Huge\boxed{\normalsize\color{lightcyan}v_{\small i}=\frac{1}{\cos\theta}\sqrt{\frac{0.5\cdot 9.8\cdot R}{R\tan\theta+H}}}$ $\text{Trajectory equation: }\;\;\Huge\boxed{\normalsize\color{lightcyan}y=R\tan\theta-\frac{9.8\cdot R^2}{v_{\small i}^2\cdot\cos^2\theta}}$ $\newline$ --- <iframe src="https://app.calconic.com/api/embed/calculator/6483e9d028c7470029d3d664" sandbox="allow-same-origin allow-forms allow-scripts allow-top-navigation allow-popups-to-escape-sandbox allow-popups" title="Calconic_ Calculator" name="Calconic_ Calculator" height="7000" scrolling="no" style="width: 100%; border: 0; outline: none;"></iframe>