Solution
I was hoping for a solution that avoided calculus and was also somehow intuitive. After all, the general answer is very simple: split the difference between the ramp angle and the vertical. Several of you sent in solutions that avoid calculus: with some trig manipulations, one can simplify the objective function so that its maximum is self-evident. Here is a concise argument by Patrick Fitzsimmons, Univ. of Calif., San Diego.
If the elevation angle of the cannon is B, then use it in the standard formula for the height of a projectile under the influence of gravity:
s(t) = v0 * t − (g/2) * t²
Recall the trigonometry identity:
sin(x)cos(y) = [sin(x + y) + sin(x − y)]/2)
A little more trigonometry shows that the horizontal distance traveled by the shirt before landing in the stands is
[ v² / g * cos A ] * [ sin(2B − A) − sin A ]
for -pi/2 < A < B < pi/2.
Here, v is the speed of the t-shirt as it leaves the cannon, and g is acceleration due to gravity.
This is clearly maximized when 2B − A = pi/2; that is, when B = A/2 + pi/4 = A + 1/2( pi/2 − A).
My hope for a more intuitive solution was not completely in vain, though. The result can be derived in a way that makes no use of calculus or trigonometry!
Richard Guy pointed me to an old book on parabolas that essentially does it, and then John Sullivan worked out the full details in an elegant way. In fact, as you will see if you download these notes, we have FOUR approaches to the problem, all of which are basically geometrical. And regardless of whether you prefer the calculus proof you know, Fitzsimmons' proof above, or the geometry proof in our notes, it is clear that the proof in our notes is giving more information: in particular, a description of the envelope of all trajectories, which turns out to be a vertically oriented parabola with focus at the origin and directrix twice as high as the (common) directrix to all the trajectories.
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© Copyright 2016 Stan Wagon. Reproduced with permission.