Page History

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

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.

\begin{large}\[t = \mbox{0 s}\]\[x =\mbox{0 m}\]\[y=\mbox{0 m}\]\end{large}

\begin{large}\[y=\mbox{1.22 m}\]\[v_{y} = \mbox{0 m/s}\]\end{large}

\begin{large}\[x = \mbox{1.83 m}\]\[y=\mbox{0 m}\]\end{large}

One way that we can solve the problem is to separate the vertical and horizontal motions and treat them independently, as functions of time..

We first analyze the motion from the launch point up to max height. For this portion of the motion, we can summarize our givens:

\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 v_y,i, since the problem is asking us for the initial launch velocity. The most direct approach is to use:

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

which becomes:

\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}

We choose the positive sign, since clearly the stream is moving upward at the instant of launch. Thus,

\begin{large} \[ v_{y,{\rm i}} = + \mbox{4.9 m/s}\]\end{large}

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:

\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}

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:

\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}

this is simplified by substituting t_i = 0 and y_i = 0:

\begin{large}\[ 0 = v_{y,{\rm i}} t + \frac{1}{2} a_{y} t^{2} \]\end{large}

This reduced version can be solved without appealing to the quadratic equation (simply factor out a t):

\begin{large}\[ t = \mbox{0 s}\qquad\mbox{or}\qquad t = -\frac{2v_{y,{\rm i}}}{a_{y}} = \mbox{1.0 s} \] \end{large}

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:

\begin{large}\[ x = x_{\rm i} + v_{x} (t-t_{\rm i}) \]\end{large}

meaning:

 \begin{large} \[ v_{x} = \frac{x}{t} = -\frac{x a_{y}}{2v_{y,{\rm i}}} = \mbox{1.8 m/s} \] \end{large}

We are not finished yet, since we are asked for the complete initial velocity. The magnitude of the full velocity is

\begin{large} \[ v_{\rm i} = \sqrt{v_{y,{\rm i}}^{2} + v_{x}^{2}} = \mbox{5.2 m/s} \] \end{large}

which allows us to draw the complete vector triangle:

and to find the angle

\begin{large} \[ \theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}}\right) = 70^{\circ} \] \end{large}

so the velocity should be 5.2 m/s at 70° above the horizontal.