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\author{BeeBooBopNerd}
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  /Title (physics-projectile-motion.pdf)
  /Creator (Cheatography)
  /Author (BeeBooBopNerd)
  /Subject (Physics - Projectile Motion Cheat Sheet)
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    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Physics - Projectile Motion Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{BeeBooBopNerd} via \textcolor{DarkBackground}{\uline{cheatography.com/131975/cs/26636/}}}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}BeeBooBopNerd \\
  \uline{cheatography.com/beeboobopnerd} \\
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 17th February, 2021.\\
   Updated 17th February, 2021.\\
   Page {\thepage} of \pageref{LastPage}.
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  Measure your website readability!\\
  www.readability-score.com
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\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Vocabulary}}  \tn
% Row 0
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{\bf{projectile}}\{\{noshy\}\}\{\{bt\}\} & \{\{bt\}\}object moving through the air, either initially thrown or dropped, subject only to the effects of gravity \tn 
% Row Count 6 (+ 6)
% Row 1
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{\bf{tragectory}}\{\{noshy\}\} & the path of a projectile, which is parabolic in two dimensions \tn 
% Row Count 9 (+ 3)
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{\bf{projectile motion}}\{\{noshy\}\} & movement of an object through the air, subject only to the effects of gravity \tn 
% Row Count 13 (+ 4)
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{\bf{range}} & the maximum horizontal distance a projectile travels \tn 
% Row Count 16 (+ 3)
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{\bf{launch angle}}\{\{noshy\}\} & The angle of a projectile's initial velocity when measured from the horizontal direction. These angles are typically 90° or less \tn 
% Row Count 22 (+ 6)
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Kinematic Equations}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Common Mistakes and Misconceptions}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{`1.) {\bf{Remember: What happens in the vertical direction does NOT affect the horizontal direction, and vise versa.}}` \newline % Row Count 3 (+ 3)
`- An object's horizontal position, velocity, or acceleration does not affect it's vertical position, velocity, or acceleration. These variables are only related by {\emph{t}} time.` \newline % Row Count 7 (+ 4)
`2.) {\bf{It's easy to forget that horizontal motion has constant velocity (and zero acceleration) while vertical motion has constant acceleration}}` \newline % Row Count 10 (+ 3)
`- This means for projectile motion, the initial velocity in the x-direction will be the same as the final velocity in the x-direction, while the starting and end velocities in the y-direction will be different because of acceleration due to gravity.` \newline % Row Count 16 (+ 6)
`3.) {\bf{Make sure to define the coordinate axes and pay attention to the sign of the acceleration constant {\emph{g}}.}}`% Row Count 19 (+ 3)
} \tn 
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{How to Solve (Launched at an Angle)}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{`{\bf{1.) Draw a diagram of the scenario}}` \newline % Row Count 1 (+ 1)
`- Make sure to label everything or Brian will be mad` \newline % Row Count 3 (+ 2)
`{\bf{2.) List our known and unknown variables}}` \newline % Row Count 4 (+ 1)
`- Make a T-chart with an {\emph{x}} and {\emph{y}} column where you fill out the variables` \newline % Row Count 6 (+ 2)
`{\bf{3.) Break the motion into horizontal and vertical components parallel to the x and y-axes}}` \newline % Row Count 8 (+ 2)
`- Motion in each dimension is independent of each other` \newline % Row Count 10 (+ 2)
`{\bf{4.) Solve for the unknowns in two separate motions - one horizontal and one vertical.}}` \newline % Row Count 12 (+ 2)
`- Use the kinematic equations to solve. Usually, try to find time first because that will make everything easier. Time is the common variable between the {\emph{x}} motion and {\emph{y}} motion` \newline % Row Count 16 (+ 4)
{\bf{When solving for the initial velocities, you have to use trig, so x would be the initial velocity times cosθ and y would be the initial velocity times sinθ}}% Row Count 20 (+ 4)
} \tn 
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{How to solve (Horizontal Projectiles)}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{`{\bf{1.) List our known and unknown variables}}` \newline % Row Count 1 (+ 1)
`- make a t-chart with an x and y column where you fill out the variables` \newline % Row Count 3 (+ 2)
`{\bf{2.) Break the motion into horizontal and vertical components parallel to the x and y-axes}}` \newline % Row Count 5 (+ 2)
`- Motion in each dimension is independent of each other` \newline % Row Count 7 (+ 2)
`{\bf{3.) Solve for the unknowns in two separate motions - one horizontal and one vertical.}}` \newline % Row Count 9 (+ 2)
`- Use the kinematic equations to solve. Usually try to find time first because that will make everything easier. Time is the common variable between the x motion and y motion`% Row Count 13 (+ 4)
} \tn 
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Tips (Horizontal Projectiles)}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{`- Um just make sure to always find time first because that makes everything a lot easier. Usually, if you want to find time, the equation is D=Vi{\emph{t}}+(1/2){\emph{a}}{\emph{t}}` \newline % Row Count 4 (+ 4)
`- Also, you usually know the initial and final velocities for the x-axis, so write that in the T chart. They should both be the same, so that means acceleration is 0. If there are any other variables that are given, write them in the T chart. As for the y-axis, acceleration should be -9.8 m/s\textasciicircum{}2\textasciicircum{}. Again, time is the same for both x and y (unless it's some weird problem)` \newline % Row Count 12 (+ 8)
`- The final velocity for y, when it reaches the vertex, is 0 m/s\textasciicircum{}2`% Row Count 14 (+ 2)
} \tn 
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% That's all folks
\end{multicols*}

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