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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
System:
Interactions:
Model:
Approach:
Understand the 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 |
|---|---|---|
Unknown macro: {latex} \begin Unknown macro: {large} [t = \mbox Unknown macro: {0 s}
][x =\mbox Unknown macro: {0 m}
][y=\mbox ]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [y=\mbox Unknown macro: {1.22 m}
][v_ Unknown macro: {y}
= \mbox Unknown macro: {0 m/s}
]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [x = \mbox Unknown macro: {1.83 m}
][y=\mbox Unknown macro: {0 m}
]\end |
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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..
Decompose the Problem
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
[t_
= \mbox
] [ x_
= \mbox
][y_
= \mbox
][y=\mbox
][v_
= \mbox
][a_
= -\mbox
^
]\end
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:
\begin
[ v_
^
= v_{y,{\rm i}}^
+ 2 a_
(y-y_
) ] \end
which becomes:
\begin
[ v_{y,{\rm i}} = \pm \sqrt{-2 a_
y} = \pm \sqrt{2 (\mbox
^
)(\mbox
)}
= \pm \mbox
]\end
We choose the positive sign, since clearly the stream is moving upward at the instant of launch. Thus,
\begin
[ v_{y,{\rm i}} = + \mbox
]\end
Reassemble the Problem
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
[t_
= \mbox
] [ x_
= \mbox
] [ x = \mbox
][y_
= \mbox
][y = \mbox
] [v_{y,{\rm i}} = \mbox
] [a_
= -\mbox
^
]\end
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:
\begin
[ y(t) = y_
+ v_{y,{\rm i}}(t-t_
) + \frac
a_
(t-t_
)^
]\end
this is simplified by substituting ti = 0 and yi = 0:
\begin
[ 0 = v_{y,{\rm i}} t + \frac
a_
t^
]\end
This reduced version can be solved without appealing to the quadratic equation (simply factor out a t):
\begin
[ t = \mbox
\qquad\mbox
\qquad t = -\frac{2v_{y,
}}{a_{y}} = \mbox
] \end
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:
\begin
[ x = x_
+ v_
(t-t_
) ]\end
meaning:
\begin
[ v_
= \frac
= -\frac{x a_{y}}{2v_{y,
}} = \mbox
] \end
We are not finished yet, since we are asked for the complete initial velocity. The magnitude of the full velocity is
\begin
[ v_
= \sqrt{v_{y,{\rm i}}^
+ v_
^{2}} = \mbox
] \end
which allows us to draw the complete vector triangle:
and to find the angle
\begin
[ \theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}}\right) = 70^
] \end
so the velocity should be 5.2 m/s at 70° above the horizontal.