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Excerpt
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Perhaps this parasitic plant should be called "Dwarf Missiletoe".
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(Photo by the Forest Service of the U.S. Department of Agriculture.)
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According to the U.S. Forest Service1, dwarf mistletoe is a parasitic plant that grows on the branches of pine trees. The mistletoe extracts its water and nutrients directly from the tree. One rare aspect of dwarf mistletoe is its seed dispersal mechanism. Rather than relying on birds or wind to spread seeds from pine tree to pine tree, mature mistletoe fruit literally explodes (as a result of extreme water pressure within the fruit). The explosion hurles the seed away from the pine tree. The seeds are coated with a sticky substance which causes them to adhere to whatever they hit. Ideally, the seed hits another nearby pine tree and begins to sprout.
The seed dispersal mechanism has been studied by T.E. Hinds and F.G. Hawksworth2 using high-speed photography. They find that Arceuthobium cyanocarpum (the variety shown in the picture above) ejects its seeds with a speed of about 2100 cm/s.
Suppose that a certain dwarf mistletoe fruit expels a seed with a velocity of 2100 cm/s directed at 30° above the horizontal. Suppose further that the seed hits another tree at exactly the same height that it was launched from. Neglecting air resistance (note: neglecting air resistance is a poor assumption in this case) how far horizontally is the landing site displaced from the launch site?
Constant external force from the earth's (gravity (near-earth)).
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Models:
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Projectile motion, assuming One-Dimensional Motion with Constant Velocity in the horizontal direction and One-Dimensional Motion with Constant Acceleration in the vertical direction.
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{excerpt:hidden=true}Perhaps this parasitic plant should be called "Dwarf Missiletoe".{excerpt}
| !dwarfmistletoe.jpg! |
| (Photo by the Forest Service of the U.S. Department of Agriculture.) |
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According to the U.S. Forest Service{^}1^, dwarf mistletoe is a parasitic plant that grows on the branches of pine trees. The mistletoe extracts its water and nutrients directly from the tree. One rare aspect of dwarf mistletoe is its seed dispersal mechanism. Rather than relying on birds or wind to spread seeds from pine tree to pine tree, mature mistletoe fruit literally explodes (as a result of extreme water pressure within the fruit). The explosion hurles the seed away from the pine tree. The seeds are coated with a sticky substance which causes them to adhere to whatever they hit. Ideally, the seed hits another nearby pine tree and begins to sprout.
The seed dispersal mechanism has been studied by T.E. Hinds and F.G. Hawksworth{^}2^ using high-speed photography. They find that _Arceuthobium cyanocarpum_ (the variety shown in the picture above) ejects its seeds with a speed of about 2100 cm/s.
Suppose that a certain dwarf mistletoe fruit expels a seed with a velocity of 2100 cm/s directed at 30° above the horizontal. Suppose further that the seed hits another tree at exactly the same height that it was launched from. Neglecting air resistance (*note:* neglecting air resistance is a poor assumption in this case) how far horizontally is the landing site displaced from the launch site?
^1^ Taylor, Jane E. and Mathiason, Robert L. ["Limber Pine Dwarf Mistletoe", _Forest Insect and Disease Leaflet_ 171|http://www.fs.fed.us/r6/nr/fid/fidls/f171.htm]. U.S. Dept. of Agriculture, Forest Service, 1999.
^2^ _Science_, Vol. 148, No. 3669 (Apr. 23 1965), pp. 517-519
h4. Solution
{toggle-cloak:id=sys} *System:* {cloak:id=sys}The seed is treated as a [point particle].{cloak}
{toggle-cloak:id=int} *Interactions:* {cloak:id=int}Constant external force from the earth's ([gravity (near-earth)|gravity (near-earth)]).{cloak}
{toggle-cloak:id=mod} *Models:* {cloak:id=mod}Projectile motion, assuming [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:*
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{toggle-cloak:id=diag} {color:red}{*}Diagrammatic Representation{*}{color}
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We first sketch the situation and define a coordinate system with acceleration only along the (negative) y-axis..
!mistletoetraj.jpg!
We choose the launch point to have the coordinates _x_ = 0 m, _y_ = 0 m and we choose to make _t_ = 0 s at the instant of launch. We are now almost ready to summarize our givens, but we first have to deal with the velocity. When putting the information given in the problem into a form suitable for use in our equations, we must break up all vector quantities into their _x_ and _y_ components. For the initial velocity, this is done by constructing a vector triangle:
!mistletoecomp.jpg!
Using this triangle, we can see that:
{latex}\begin{large}\[ v_{x} = v \cos \theta = \mbox{18.2 m/s} \]\[ v_{y,{\rm i}} = v \sin \theta = \mbox{10.5 m/s}\]\end{large}{latex}
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{toggle-cloak:id=math} {color:red}{*}Mathematical Representation{*}{color}
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We can now state our givens:
{panel:title=givens}{latex}\begin{large}\[ t_{\rm i} = \mbox{0 s}\]\[x_{\rm i} = \mbox{0 m}\] \[y_{\rm i} = \mbox{0 m}\]\[y = \mbox{0 m} \]\[ v_{x} = \mbox{18.2 m/s} \] \[ v_{y,{\rm i}} = \mbox{10.5 m/s} \]\[a_{y} = -\mbox{9.8 m/s}^{2}\]\end{large}{latex}{panel}
Our end goal is to determine _x_, which will tell us the horizontal displacement of the seed during its flight. As usual, however, we must first find the time by using the _y_ direction. The most direct approach is to use the Law of Change to find the time when the projectile returns to height y=0:
{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, after substituting {latex}$y(t_R) = 0${latex}, can be solved to give:
{latex} \begin{large} \[ t_{R} = \mbox{0 s} \qquad \mbox{or}\qquad t_R = -\frac{2 v_{y}}{a_{y}} = \mbox{2.14 s} \] \end{large}{latex}
We can now solve the _x_ direction Law of Change:
{latex}\begin{large} \[ x(t_R) = x_{\rm i} + v_{x} (t-t_{rm i}) = -\frac{2 v_{x} v_{y}}{a_{y}} = \mbox{39 m} \] \end{large}{latex}
{tip}A good check is to think about whether this answer is _reasonable_. Given what you know about pine trees, does a 39 m range seem large enough to make it likely the seed will hit another tree?{tip}
{info}This problem is one where air resistance makes much more than a 10% difference in the answer. The actual range of these dwarf mistletoe seeds is estimated by the Forest Service to be about 40 *feet* rather than 40 meters. But on a windy day, this could change by a factor of 2.
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h4. Follow Up
{toggle-cloak:id=follow} *The Range Formula*
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Suppose that instead of calculating _v{_}{_}{~}x{~}_ and _v{_}{_}{~}y{~}_ explicitly to find our givens, we had just substituted the expressions _v_ cosθ for _v{_}{_}{~}x{~}_ and _v_ sinθ for _v{_}{_}{~}y{~}_. You can check that the answer we found could have been written:
{latex}\begin{large} \[ x = -\frac{2 v^{2} \cos\theta \: \sin\theta}{a_{y}} = \frac{2 v^{2} \cos\theta \: \sin\theta}{g} \]\end{large}{latex}
By using the trig identities for the sine of a sum, you can show that this becomes:
{latex}\begin{large} \[ x = \frac{v^{2} \sin(2\theta)}{g} \] \end{large}{latex}
This simple result is sometimes called the _range formula_. Remember how it is obtained: find the time until the projectile hits the ground from the y-motion; then find the x-position at that time.
{warning}The formula derived here only gives the horizontal range accurately if the projectile lands at exactly the same height it was launched from (and if air resistance is negligible). It is a common mistake to use this formula in cases where the launch height and landing height are different.{warning}
You can use this range formula to prove two standard claims about projectiles fired over level ground:
# The range of projectiles fired with the same speed at complimentary angles (θ and 90°-θ) will be the same.
# The range of a projectile will be maximized when the launch angle is 45°, assuming the launch speed is independent of angle.
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