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{
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 Excerpt
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Modern 
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 fountains 
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 are 
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 an 
...
 excellent 
...
 example 
...
 of 
...
 projectile 
...
motion. 
 Image Added 
 Photo courtesy Wikimedia Commons. 
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?
Solution
 Toggle Cloak
idsys
 System: 
 Cloak
idsys
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 .
 Toggle Cloak
idint
 Interactions: 
 Cloak
idint
The system is in freefall (influenced only by the earth's gravity).
 Toggle Cloak
idmod
 Model: 
 Cloak
idmod
Projectile Motion : One-Dimensional Motion with Constant Velocity in the horizontal direction and One-Dimensional Motion with Constant Acceleration in the vertical direction.
 Toggle Cloak
idapp
 Approach:
 Cloak
idapp
 Toggle Cloak
idgivens
 Understand the Givens
 Cloak
idgivens
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
\begin{large}\[y=\mbox{1.22 m}\]\[v_{y} = \mbox{0 m/s}\]\end{large}
...
 Latex
\begin{large}\[x = \mbox{1.83 m}\]\[y=\mbox{0 m}\]\end{large}
Image Added
  
 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). 
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
givens
 Toggle Cloak
idpart1
 Decompose the Problem
 Cloak
idpart1
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
givens for upward motion
{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}
 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}
We would like to solve for vy,i, since the problem is asking us for the initial launch velocity. The most direct approach is to use:
 Latex
{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 
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 positive 
...
 sign, 
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 since 
...
 clearly 
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 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}
{
 Cloak
part1
part1
 Toggle Cloak
...
id
...
part2
 Reassemble the Problem
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idpart2
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
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}
 Info
Note that it is the fact that both y and yi are 0 m for the full trajectory which forced us to first consider the upward portion. 
We would like to find vx, 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
{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}\[ y(t) = y_{\rm i} + v_{y,{\rm i}}(t-t_{\rm i}) + \frac{1}{2}a_{y}(t-t_{\rm i})^{2} \]\end{large}{latex}

which is greatly simplified after inserting zeros:

{latex}
this is simplified by substituting ti = 0 and yi = 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 
...
 vx:
 Latex
_~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:
Image Added
and to find the angle
 Latex


!velvec.png!

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.
 at 70° above the horizontal.

{cloak:part2}
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part2
part2