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{excerpt:hidden=true}Modern fountains are an excellent example of projectile motion. {excerpt} 

| !fountain.jpg|width=650! |
| Photo courtesy [Wikimedia Commons|http://commons.wikimedia.org]. |

Suppose you are designing a fountain that will shoot jets of water. The water jets will emerge from nozzles at the same level as the pool they fall into. If you want the jets to reach a height of *4.0 feet* above the water's surface and to travel *6.0 feet* horizontally (ignoring air resistance), with what velocity should the water leave the nozzles?

h4. Solution

{toggle-cloak:id=sys} *System:* {cloak:id=sys}We will imagine breaking the stream of water up into small pieces (think of water droplets making up the stream). We will examine the trajectory of _one_ of the pieces, treating it as a [point particle].{cloak}

{toggle-cloak:id=int} *Interactions:* {cloak:id=int}The system is in freefall (influenced only by the earth's gravity).
{cloak}

{toggle-cloak:id=mod} *Model:* {cloak:id=mod}Projectile Motion : [One-Dimensional Motion with Constant Velocity|1-D Motion (Constant Velocity)] in the horizontal direction and [One-Dimensional Motion with Constant Acceleration|1-D Motion (Constant Acceleration)] in the vertical direction.{cloak}

{toggle-cloak:id=app} *Approach:*
{cloak:id=app}
{toggle-cloak:id=givens} {color:red}{*}Understand the Givens{*}{color}
{cloak:id=givens}
We choose a coordinate system where the stream travels in the {*}+ x{*} direction and the {*}+ y{*} direction points upward. Further, we choose the surface of the fountain pool to be at the level {*}y = 0 m{*} and the nozzle to be at the point {*}x = 0 m{*}. We also choose {*}t = 0 s{*} at the instant of launch from the nozzle. We immediately run into a problem, however. The difficulty here is that we have information about three separate points. 

||Launch||Max Height||Landing||
|{latex}\begin{large}\[t = \mbox{0 s}\]\[x =\mbox{0 m}\]\[y=\mbox{0 m}\]\end{large}{latex}|{latex}\begin{large}\[y=\mbox{1.22 m}\]\[v_{y} = \mbox{0 m/s}\]\end{large}{latex}|{latex}\begin{large}\[x = \mbox{1.83 m}\]\[y=\mbox{0 m}\]\end{large}{latex}

 |!Fountain Drawing.png!|   

{note}We use the fact that the vertical velocity goes to zero when the water reaches the maximum height to analyze one-dimensional freefall. You can see that this is still true for two-dimensional projectile motion by making a plot of *y* versus time. Note that the slope goes to zero (the curve is horizontal) at the maximum height. It is important to remember, however, that the *x* velocity is *not* zero at any point in 2-D projectile motion (it is a constant). {note}
One way that we can solve the problem is to separate the vertical and horizontal motions and treat them independently, as functions of time..
{cloak:givens}
{toggle-cloak:id=part1} {color:red}{*}Decompose the Problem{*}{color}
{cloak:id=part1}
We first analyze the motion from the launch point up to max height. For this portion of the motion, we can summarize our givens:
{panel:givens for upward motion}
{latex}\begin{large}\[t_{\rm i} = \mbox{0 s}\] \[ x_{\rm i} = \mbox{0 m}\]\[y_{\rm i} = \mbox{0 m} \]\[y=\mbox{1.22 m}\]\[v_{y} = \mbox{0 m/s} \]\[a_{y} = -\mbox{9.8 m/s}^{2}\]\end{large}{latex}
{panel}
We would like to solve for {*}v{~}y,i{~}{*}, since the problem is asking us for the initial launch velocity. The most direct approach is to use:
{latex}\begin{large}\[ v_{y}^{2} = v_{y,{\rm i}}^{2} + 2 a_{y}(y-y_{\rm i}) \] \end{large}{latex}
which becomes:
{latex}\begin{large} \[ v_{y,{\rm i}} = \pm \sqrt{-2 a_{y} y} = \pm \sqrt{-2 (-\mbox{9.8 m/s}^{2})(\mbox{1.22 m})}
= \pm \mbox{4.9 m/s} \]\end{large}{latex}
We choose the positive sign, since clearly the stream is moving upward at the instant of launch. Thus,
{latex}\begin{large} \[ v_{y,{\rm i}} = + \mbox{4.9 m/s}\]\end{large}{latex}
{cloak:part1}
{toggle-cloak:id=part2} {color:red}{*}Reassemble the Problem{*}{color}
{cloak:id=part2}
Now we have to find the *x* velocity. The most direct way to do this is to now consider the entire motion as one part. If we take the whole trajectory, we have the givens:
{panel:givens for full trajectory}
{latex}\begin{large}\[t_{\rm i} = \mbox{0 s}\] \[ x_{\rm i} = \mbox{0 m} \] \[ x = \mbox{1.83 m} \]\[y_{\rm i} = \mbox{0 m}\]\[y = \mbox{0 m} \] \[v_{y,{\rm i}} = \mbox{4.9 m/s} \] \[a_{y} = -\mbox{9.8 m/s}^{2}\]\end{large}{latex}
{panel}{info}Note that it is the fact that both *y* and {*}y{~}i{~}{*} are {*}0 m{*} for the full trajectory which forced us to first consider the upward portion. {info}
We would like to find {*}v{~}x~{*}, but we must first solve for the time by using the equation of motion in the *y* direction. The most direct way to obtain the time is to use:
{latex}\begin{large}\[ y(t) = y_{\rm i} + v_{y,{\rm i}}(t-t_{\rm i}) + \frac{1}{2}a_{y}(t-t_{\rm i})^{2} \]\end{large}{latex}
this is simplified by substituting {*}t{~}i{~} = 0{*} and {*}y{~}i{~} = 0{*}:
{latex}\begin{large}\[ 0 = v_{y,{\rm i}} t + \frac{1}{2} a_{y} t^{2} \]\end{large}{latex}
This reduced version can be solved without appealing to the quadratic equation (simply factor out a *t*):
{latex}\begin{large}\[ t = \mbox{0 s}\qquad\mbox{or}\qquad t = -\frac{2v_{y,{\rm i}}}{a_{y}} = \mbox{1.0 s} \] \end{large}{latex}
We can rule out the {*}t = 0 s{*} solution, since that is simply reminding us that the water was launched from the level of the pool at {*}t = 0 s{*}. The water will return to the level of the pool {*}1.0 s{*} after launch. With this information, we can solve for {*}v{~}x~{*}:
{latex}\begin{large}\[ x = x_{\rm i} + v_{x} (t-t_{\rm i}) \]\end{large}{latex}
meaning:
{latex} \begin{large} \[ v_{x} = \frac{x}{t} = -\frac{x a_{y}}{2v_{y,{\rm i}}} = \mbox{1.8 m/s} \] \end{large}{latex}
We are not finished yet, since we are asked for the complete initial velocity. The magnitude of the full velocity is
{latex}\begin{large} \[ v_{\rm i} = \sqrt{v_{y,{\rm i}}^{2} + v_{x}^{2}} = \mbox{5.2 m/s} \] \end{large}{latex}
which allows us to draw the complete vector triangle:

!velvec.jpg!

and to find the angle
{latex}\begin{large} \[ \theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}}\right) = 70^{\circ} \] \end{large}{latex}
so the velocity should be 5.2 m/s at 70° above the horizontal.
{cloak:part2}

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