Chapter 6: Basic Physics SEProjectile Motion
The Big Idea
In this chapter, we aim to understand and explain the parabolic motion of a thrown object, known as projectile motion. Motion in one direction is unrelated to motion in other perpendicular directions. Once the object has been thrown, the only acceleration is in the \begin{align*}y\end{align*}
Key Equations
In the vertical direction

\begin{align*}\Delta y = v_{iy} t  \frac{1}{2} gt^2\end{align*}
Δy=viyt−12gt2 
\begin{align*}\Delta v_y =  gt\end{align*}
Δvy=−gt 
\begin{align*}v_{y}{^2} = v_{iy}{^2}  2g(\Delta y)\end{align*}
vy2=viy2−2g(Δy) 
\begin{align*}a_y = g = 9.806 \ m/s^2 \approx 10 \ m/s^2\end{align*}
ay=−g=−9.806 m/s2≈−10 m/s2
In the horizontal direction

\begin{align*}\Delta x = v_{ix}t\end{align*}
Δx=vixt 
\begin{align*}v_x = v_{ix}\end{align*}
vx=vix (does not change) 
\begin{align*}a_x = 0\end{align*}
ax=0
A few comments on the above equations:
 Recall: The ‘\begin{align*}\Delta\end{align*}
Δ ’ symbol means ‘change in’ so that for example \begin{align*}\Delta v_y = v_{fy}  v_{iy}\end{align*}Δvy=vfy−viy , which tells you the change in the velocity from the starting velocity to the final velocity. 
\begin{align*}+y\end{align*}
+y direction is defined as upward for the above equations  The initial velocity \begin{align*}v_i\end{align*}
vi can be separated into \begin{align*}v_{ix} = v_i \cos \theta\end{align*}vix=vicosθ and \begin{align*}v_{iy} = v_i \sin \theta\end{align*}viy=visinθ , where \begin{align*}\theta\end{align*}θ is the angle between the velocity vector and the horizontal.
Key Concepts
 In projectile motion, the horizontal displacement of an object is called its range.
 At the top of its flight, the vertical speed of an object is zero.
 To work these problems, use the equations above: one set for the \begin{align*}y\end{align*}
y− direction (vertical direction), and one set for the \begin{align*}x\end{align*}x− direction (horizontal direction). The \begin{align*}x\end{align*}x− direction and \begin{align*}y\end{align*}y− direction don’t “talk” to each other. They are separate dimensions. Keep them separate.  The time is the same for the two directions, and can be plugged into both equations.
 Since in the absence of air resistance, there is no acceleration in the \begin{align*}x\end{align*}
x− direction, the velocity in the \begin{align*}x\end{align*}x− direction does not change over time. This is a counterintuitive notion for many. (Air resistance will cause the \begin{align*}x\end{align*}x− velocity to decrease slightly or significantly depending on the object. But this factor is ignored for the time being.)  The \begin{align*}y\end{align*}
y− direction motion must include the acceleration due to gravity, and therefore the velocity in the \begin{align*}y\end{align*}y− direction changes over time.  The shape of the path of an object undergoing projectile motion is a parabola.
 We will ignore air resistance in this chapter. Air resistance will tend to shorten the range of the projectile motion by virtue of producing an acceleration opposite to the direction of motion.
Solved Examples
Example 1: A tennis ball is launched \begin{align*}32^\circ\end{align*}
Question: \begin{align*}v_x\end{align*}
Given: \begin{align*}v = 7.0 \ m/s\end{align*}
\begin{align*}{\;} \qquad \quad \theta = 32^\circ\end{align*}
Equation: \begin{align*}v_x = v \cos \theta \qquad v_y = v \sin \theta\end{align*}
Plug n’ Chug: \begin{align*}v_x = v \cos \theta = (7.0 \ m/s) \cos (32^\circ)=5.9 \ m/s\end{align*}
\begin{align*}{\;}\qquad \qquad \qquad \ v_y = v \sin \theta = (7.0 \ m/s) \sin (32^\circ)= 3.7 \ m/s\end{align*}
Answer: \begin{align*}\boxed{\mathbf{5.9 \ m/s, 3.7 \ m/s.}}\end{align*}
Example 2: CSI discovers a car at the bottom of a 72 m cliff. How fast was the car going if it landed 22m horizontally from the cliff’s edge? (Note that the cliff is flat, i.e. the car came off the cliff horizontally).
Question: \begin{align*}v = \ ? \ [m/s]\end{align*}
Given: \begin{align*}h = \Delta y = 72 \ m\end{align*}
\begin{align*}{\;}\qquad \quad d = \Delta x = 22 \ m\end{align*}
\begin{align*}{\;}\qquad \quad g = 10.0 \ m/s^2\end{align*}
Equation: \begin{align*}h = v_{iy} t + \frac{1}{2} gt^2\end{align*} and \begin{align*}d = v_{ix} t\end{align*}
Plug n’ Chug: Step 1: Calculate the time required for the car to freefall from a height of 72 m.
\begin{align*}h = v_{iy} t + \frac{1}{2} gt^2\end{align*} but since \begin{align*}v_{iy}=0\end{align*}, the equation simplifies to \begin{align*}h = \frac{1}{2} gt^2\end{align*} rearranging for the unknown variable, \begin{align*}t\end{align*}, yields
\begin{align*}t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(72 \ m)}{10.0 \ m/s^2}} = 3.79 \ s\end{align*}
Step 2: Solve for initial velocity:
\begin{align*}v_{ix} = \frac{d}{t} = \frac{22 \ m}{3.79 \ s} = 5.80 \ m/s\end{align*}
Answer: \begin{align*}\boxed{\mathbf{5.80 \ m/s}}\end{align*}
TwoDimensional and Projectile Motion Problem Set
Draw detailed pictures for each problem (putting in all the data, such as initial velocity, time, etc.), and write down your questions when you get stuck.