letters to the editor
Analytically solving Tarzan’s dilemma
Teachers who use the instructive Tarzan exercise in the
November 2013 issue1 may find it useful to know that TPT
previously published an analytic solution2 for the maximum
range. To review the problem, Tarzan grabs the end of a vertically hanging vine of length L at height h above the ground
with initial speed v­. He swings up to angle q and then lets go
of the vine, landing on the ground a horizontal distance R
away from the point at which he grabbed the vine. Normalizing the variables by defining w  v2/2gL and y  h/L, then
the angle q max at which Tarzan should release the vine to
maximize his range is given by
(1)
where x  cos qmax. Equation (1) is a cubic polynomial
in x with one real solution that can be straightforwardly
obtained using Cardano’s formula (but which is a bit long to
write down here). For example, using the values v = 10 m/s,
h = 1 m, L = 3 m, and g = 9.8 m/s2, the solution is
qmax = 40.3°, which is more exact than the numerical value
39.4° estimated in Ref. 1. Two other recent articles3,4 also
discuss interesting aspects of this problem.
References
1.
2.
3.
4.
M. Rave and M. Sayers, “Tarzan’s dilemma: A challenging
problem for introductory physics students,” Phys. Teach. 51,
456–458 (Nov. 2013).
D. Bittel, “Maximizing the range of a projectile launched by a
simple pendulum,” Phys. Teach. 43, 98–100 (Feb. 2005).
C. E. Mungan and T. C. Lipscombe, “Swinging over the water
hole,” Lat. Am. J. Phys. Educ. 5, 335–337 (June 2011).
H. Shima, “How far can Tarzan jump?” Eur. J. Phys. 33, 1687–
1693 (Nov. 2012).
Carl E. Mungan
U.S. Naval Academy
Annapolis, MD 21402
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The Physics Teacher ◆ Vol. 52, January 2014