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

|!fountain.jpg!|
|(Image courtesy Wikimedia Commons.)|

{composition-setup}{composition-setup}

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}[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.  

{table}{tr}{th:align=center|bgcolor=#F2F2F2}Launch{th}{th:align=center|bgcolor=#F2F2F2}Max Height{th}{th:align=center|bgcolor=#F2F2F2}Landing{th}{tr}
{tr}{td:width=100|align=center}{latex}\begin{large}\[t = \mbox{0 s}\]\[x =\mbox{0 m}\]\[y=\mbox{0 m}\]\end{large}{latex}{td}{td:width=100|align=center}{latex}\begin{large}\[y=\mbox{1.22 m}\]\[v_{y} = \mbox{0 m/s}\]\end{large}{latex}{td}{td:width=100|align=center}{latex}\begin{large}\[x = \mbox{1.83 m}\]\[y=\mbox{0 m}\]\end{large}{latex}{td}{tr}{table}

{note}We used the fact that the vertical velocity goes to zero at the point of max height when we analyzed 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}

The only way that we can solve the problem is to break it up into two problems.  

{cloak:givens}

{toggle-cloak:id=part1} {color:red} *Divide 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 _y_ direction.  The most direct way to obtain the time is to use:

{latex}\begin{large}\[ y = y_{\rm i} + v_{y,{\rm i}}(t-t_{\rm i}) + \frac{1}{2}a_{y}(t-t_{\rm i})^{2} \]\end{large}{latex}

which is greatly simplified after inserting zeros:

{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}