According to the U.S. Forest Service ("Limber Pine Dwarf Mistletoe", Forest Insect and Disease Leaflet 171, 1999), 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 ( Science, Vol. 148, No. 3669 (Apr. 23 1965), pp. 517-519) by means of high-speed photography. They find that Arceuthobium cyanocarpum (the variety shown in the picture above) ejects is 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?
System: The seed is treated as a point particle.
Models: Projectile motion, assuming One-Dimensional Motion with Constant Velocity in the horizontal direction and One-Dimensional Motion with Constant Acceleration in the vertical direction.
Approach: We first sketch the situation and define a coordinate system.
PICTURE
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:
Using this triangle, we can see that:
\begin
[ v_
= v \cos \theta = \mbox
][ v_{y,{\rm i}} = v \sin \theta = \mbox
]\end
We can now state our givens:
\begin
[ t_
= \mbox
][x_
= \mbox
] [y_
= \mbox
][y = \mbox
][ v_
= \mbox
] [ v_{y,{\rm i}} = \mbox
][a_
= -\mbox
^
]\end
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:
\begin
[ y = y_
+ v_{y,{\rm i}} (t-t_
) + \frac
a_
(t-t_
)^
] \end
which, after substituting zeros, can be solved to give:
\begin
[ t = \mbox
\qquad \mbox
\qquad t = -\frac{2 v_{y}}{a_{y}} = \mbox
] \end
We can now solve the x direction Law of Change:
\begin
[ x = x_
+ v_
(t-t_
) = -\frac{2 v_
v_{y}}{a_{y}} = \mbox
] \end
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?
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.
Follow Up: The Range Formula
Suppose that instead of calculating vx and vy explicitly to find our givens, we had just substituted the expressions v cosθ for vx and _v_sinθ for vy. You can check that the answer we found could have been written:
\begin
[ x = -\frac{2 v^
\cos\theta \: \sin\theta}{a_{y}} = \frac{2 v^
\cos\theta \: \sin\theta}
]\end
By using the trig identities for the sine of a sum, you can show that this becomes:
\begin
[ x = \frac{v^
\sin(2\theta)}
] \end
This simple result is sometimes called the range formula.
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.
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.