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S1.3 Application: projectile motion

[A] The problem

Consider motion of projectile near the earth’s surface.

  • Choose coordinate axes:

    x - horizontal

    y - vertical

Graphic - No title - projcoords
DMy vertical may not be yours

[B] Results

(1.6)
\begin{displaymath} \mbox{{\rm $x$-motion:}} \hspace*{1cm} v_x= v_{x0} \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} x=v_{x0}t \end{displaymath}
(1.7)
\begin{displaymath} \mbox{{\rm $y$-motion:}} \hspace*{1cm} v_y= v_{y0}-g t \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} y=v_{y0}t - \frac{1}{2}gt^2 \end{displaymath}
(1.8)
\begin{displaymath} \mbox{{\rm time of flight:}} \hspace*{1cm} t_f=\frac{2v_{0}\sin \theta}{g} \end{displaymath}
(1.9)
\begin{displaymath} \mbox{{\rm range:}} \hspace*{1cm} R= \frac{2v_0^2 \sin \theta \cos \theta}{g} \end{displaymath}
(1.10)
\begin{displaymath} \mbox{{\rm trajectory equation:}} \hspace*{1cm} y= x \tan \theta -\frac{gx^2}{2v_0^2 \cos ^2 \theta} \end{displaymath}

Analysis

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Analysis

  • Formulate strategy: think separately of x and y motions.
  • The horizontal motion

    Set ax=0 and x0=0 in Key Point 1.4(a), (b)

    vx=vx0+axt  implies vx=vx0

    $x-x_0 = v_{x0} t +\frac{1}{2}a_xt^2$  implies x=vx0t

  • The vertical motion

    Set ay=-g and use the y-forms of Key Point 1.4(a) and (b):

    vy=vy0+ayt   implies vy=vy0-gt

    $y-y_0=v_{y0}t + \frac{1}{2}a_yt^2$  implies $y=v_{yo}t - \frac{1}{2}gt^2$

  • The time of flight, tf

    Flight ends on return to y=0

    Thus set

    \[ 0= y=\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{v_{y0}t - \frac{1}{2}gt^2} \]
    with solutions
    \[ \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{t=0}, \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{t= 2v_{y0}/g} \]

    First solution identifies the start (launch)

    Second solution must thus identify the end (crunch)

    Setting vy0=v0sinθ deduce

    tf=[2v0sinθ]/g

    Check It!
    • Are the units OK?
    • Does it make sense?
  • The range, R

    Range is horizontal displacement in time tf. Thus

    R=vx0tf

    Set

    vx0=v0cosθ
    and
    \[ t_f=\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{2v_{0}\sin \theta}{g}} \]

    Then

    $$R= [2v_0^2 \sin \theta \cos \theta]/g$$

    Check It!
    • Are the units OK?
    • Does it make sense?
  • The trajectory equation y(x)

    Eliminating t between

    x=vx0t

    and

    \[ y=\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{v_{y0}t - \frac{1}{2}gt^2} \]

    gives

    \[ y=v_{y0}\left[\frac{x}{v_{x0}}\right] - \frac{1}{2}g \left[\frac{x}{v_{x0}}\right] ^2 \]

    Set

    vx0=v0cosθ
    and
    vy0=v0sinθ

    $$y= x \tan \theta -[gx^2]/[ 2v_0^2 \cos ^2 \theta]$$

Check It!
  • Are the units OK?
  • Does it make sense?
 

Visualization:

Graphic - No title - projtraj
TProjectiles on Battleships
MSimulated projectiles

Learning Resources

Textbook: HRW Chapter 4.5-6
Course Questions:
Self-Test Questions: