OFFSET
0,1
COMMENTS
A projectile is launched with an initial speed v at angle theta above the horizon. Assuming that the gravitational acceleration g is uniform and neglecting the air resistance, the trajectory is a part of a parabola whose length is maximized when the angle is the root of the equation csc(theta) = coth(csc(theta)). The maximal length is then u * v^2/g, where u = 1.1996... is the root of coth(x) = x (A085984).
The angle in degrees is 56.4658351274...
The initial angle that maximizes the horizontal distance is the well-known result theta = Pi/4 = 45 degrees. The corresponding length of trajectory in this case is u * v^2/g, where u = (sqrt(2) + arcsinh(1))/2 = 1.1477... (A103711), which is 95.67...% of the maximum value.
REFERENCES
Thomas Szirtes, Applied Dimensional Analysis and Modeling, Butterworth-Heinemann, 2007, p. 578.
LINKS
Joshua Cooper and Anton Swifton, Throwing a ball as far as possible, revisited, The American Mathematical Monthly, Vol. 124, No. 10 (2017), pp. 955-959; arXiv preprint, arXiv:1611.02376 [math.HO], 2016.
Haiduke Sarafian, On projectile motion, The Physics Teacher, Vol. 37, No. 2 (1999), pp. 86-88.
Ju Yan-Qing, Projectile motion path length and initial projectile angle, Journal of Science of Teachers' College and University, Vol. 3 (2005), pp. 49-51.
FORMULA
Equals arccsc(u) where u is the root of coth(x) = x (A085984).
EXAMPLE
0.98551473786231546211492853725730463877247220596742...
MATHEMATICA
RealDigits[ArcCsc[x /. FindRoot[x == Coth[x], {x, 1}, WorkingPrecision -> 120]], 10, 100][[1]]
PROG
(PARI) solve(x=0, 1, my(s=sin(x)); s*atanh(s)-1) \\ Charles R Greathouse IV, Sep 18 2024
(PARI) asin(solve(u=.5, 1, tanh(1/u)-u)) \\ Charles R Greathouse IV, Sep 18 2024
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Amiram Eldar, Jun 25 2021
STATUS
approved