Theoretical analysis of the maximum range of a projectile released from a pendulum

Ken Yamamoto

REVIEW 2 major objections 5 minor 25 references

Theoretical analysis of the maximum range of a projectile released from a pendulum

T0 review · 2 major / 5 minor · reviewed 2026-08-11 · deepseek-v4-flash

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Pith's one-line read The optimal release angle of the Tarzan jump is the unique root of a cubic equation in cos θ, climbs with starting speed, and approaches 45 degrees.

desk verdict Clean derivation and proofs for a known cubic; two fixable gaps (an unexamined derivative branch and a sign typo) don't shake the main result. read the letter →

arxiv 2608.09157 v1 pith:OM3L2DHE submitted 2026-08-10 math-ph math.MPphysics.class-ph

Ken Yamamoto This is my paper · ORCID

classification math-phmath.MPphysics.class-ph

keywords Tarzanjumpoptimalreleaseangleprojectilemotionpendulumcubicequationasymptoticanalysismaximumrangeclassicalmechanics

verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

This paper tackles the Tarzan jump: a runner starts with horizontal speed $v_0$, swings on a rope of length $L$, and releases at angle $\theta$, landing at horizontal range $R(\theta)$. The paper's thesis is that the optimal release angle is always strictly below $45^\circ$, is the unique solution of the cubic equation $\cos^3\theta_{\max}+(\tilde H+2\tilde v_0^2-1)\cos^2\theta_{\max}-(\tilde H+\tilde v_0^2)=0$, and increases monotonically with the starting speed, approaching $45^\circ$ as $v_0\to\infty$. It proves the cubic has a unique root in the relevant interval, derives the monotonicity, and gives asymptotic formulas for both the angle and the range in the slow- and fast-start limits. A careful reader would care because a problem previously handled mostly through numerical examples here receives a complete exact analysis with clean physical limiting cases.

What carries the argument

The load-bearing machinery is a factorization of the derivative condition using two auxiliary functions, $F(\theta)=\tilde v_0^2-(1-\cos\theta)$ and $G(\theta)=\tilde H+\tilde v_0^2-F(\theta)\cos^2\theta$, which compress the range formula into $\tilde R=\sin\theta+2F\sin\theta\cos\theta+2\cos\theta\sqrt{FG}$. The pivotal identity, valid at the optimum, is $F(\theta)G(\theta)=(\tilde H+\tilde v_0^2)^2\sin^2\theta$; it eliminates the square root and, after a polynomial factorization, leaves the cubic equation in $x=\cos\theta_{\max}$. This factorization is the step that turns a messy trigonometric optimization into an algebraic root-finding problem.

What would settle it

Take a fast-start case, for instance $\tilde H=1$ and $\tilde v_0=1.5$, and numerically evaluate $\tilde R(\theta)$ over all release angles the pendulum can reach; if any stationary point from $2F(\theta)+\cos\theta=0$ beats the range predicted by the cubic root, then the claimed characterization is incomplete.

Watch

Extended reading notes

Core claim

The paper's central claim is a complete exact solution of the pendulum-release problem. The nondimensional range is written as $\tilde R(\theta)=\sin\theta+2F(\theta)\sin\theta\cos\theta+2\cos\theta\sqrt{F(\theta)G(\theta)}$, with $F(\theta)=\tilde v_0^2-(1-\cos\theta)$ and $G(\theta)=\tilde H+\tilde v_0^2-F(\theta)\cos^2\theta$, and its derivative factors as $(2F+\cos\theta)(\cos 2\theta - (G-F\cos^2\theta)\sin\theta/\sqrt{FG})$. The paper argues that the maximum comes from the second factor; squaring that factor and manipulating it produces the governing cubic $\cos^3\theta_{\max}+(\tilde H+2\tilde v_0^2-1)\cos^2\theta_{\max}-(\tilde H+\tilde v_0^2)=0$. Evaluating the cubic at $x=1$ and $x=1/\sqrt2$ and using monotonicity on $1/\sqrt2<x<1$ proves that its unique root lies there, so $0<\theta_{\max}<\pi/4$ for all $\tilde H\ge0$, $\tilde v_0>0$. Implicit differentiation then shows $\theta_{\max}$ increases monotonically with $\tilde v_0$, with asymptotic laws $\theta_{\max}\simeq\sqrt{2/(2\tilde H+1)}\,\tilde v_0$ for slow starts and $\theta_{\max}\simeq\pi/4-(\tilde H+1-1/\sqrt2)/(4\tilde v_0^2)$ for fast starts, and range limits $R\simeq v_0\sqrt{(2H+L)/g}$ and $R\simeq v_0^2/g+H+(\sqrt2-1)L$.

Load-bearing premise

The proof assumes the maximum of the range lies on the second factor of the derivative equation and never on the branch $2F(\theta)+\cos\theta=0$, a branch that is excluded only for $\tilde v_0^2<1$ and left unexamined for faster starts.

Editorial extensions

If this is right

For any rope length, drop height, and starting speed, the optimal release angle is the unique cubic root in $(0,\pi/4)$, so no numerical search is needed.
As the initial speed grows, the best release angle rises monotonically and saturates at $45^\circ$, with the gap shrinking like $v_0^{-2}$.
For slow starts the maximum range grows linearly with $v_0$; for fast starts it approaches the familiar ground-level projectile range $v_0^2/g$ plus a height correction.
When the drop height is very large compared with the rope length and speed scale, the optimal release angle collapses to $0$: a horizontal leap beats swinging on the rope.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

One testable gap the paper leaves open: for nondimensional speeds $\tilde v_0\ge1$, the first branch $2F(\theta)+\cos\theta=0$ is not fully ruled out, so a direct numerical scan of the full return map for $\tilde H=1$, $\tilde v_0=1.5$ would tell whether the cubic always gives the global maximum.
The same auxiliary-function trick could be carried over to release from other constrained paths, such as a cycloidal pendulum or a landing on an incline, where the factorization would likely yield a higher-degree polynomial rather than a cubic.
If viscous air resistance is added, the energy-conservation step that defines $F(\theta)$ fails, so the cubic cannot be exact; the asymptotic formulas here provide a zero-drag benchmark against which perturbative or numerical results for weak drag can be compared.

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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Desk Editor's Note

read the letter

Short version: this is a clean, mostly correct paper on a small problem. The cubic equation is not new—Bittel stated it—but Yamamoto supplies a genuinely efficient derivation, existence/uniqueness and monotonicity proofs, and the two asymptotic limits. The main algebra checks out; I verified the factorization and the asymptotics independently.

The good stuff: the auxiliary functions F and G make the derivative factorization transparent, the existence/uniqueness argument on (1/√2,1) is solid, and the monotonicity proof is airtight. The asymptotics match the numerics, and the paper is honest about prior work, citing Bittel, Mungan, and Rave/Sayers properly. The alternative short-v derivation in Section D is a nice sanity check.

Soft spots, all easy to patch. First, the derivation of the cubic drops the first factor 2F + cosθ = 0 after ruling it out only for ṽ₀² < 1. For ṽ₀² ≥ 1 the branch has physical solutions in the range 1 ≤ ṽ₀² ≤ 5/2, and the paper never shows they are not global maximizers. They are not: on that branch cosθ ≤ 0 and F ≥ 0, so R(θ) ≤ sinθ ≤ 1, while θ = π/4 gives R > 1 for every ṽ₀² ≥ 1. That argument should be inserted. Second, the displayed equation before the factorization has the sign of the (H̃+ṽ₀²)² sin²θ term wrong; it should be minus. The factorization corresponds to the corrected sign. Third, the squaring step from Eq. (10) to Eq. (13) is not checked for extraneous roots. The check is short: for the cubic root x = cosθmax in (1/√2,1), G − F x² = x²(H̃+1−x) > 0, so the original unsquared equation holds with the correct sign. None of this changes the final results.

Who is this for? Physics educators and anyone wanting a complete benchmark for the Tarzan jump. It does not open a new direction, but it fills a known gap with tidy proofs. I would send it to peer review with a request to add the branch argument and fix the typo; it is not a desk reject.

Referee Report

2 major / 5 minor

Summary. The paper analyzes the optimal release angle θmax and the corresponding maximum horizontal range R(θmax) for a projectile released from a simple pendulum (the 'Tarzan jump' problem). After deriving the nondimensional range function R~(θ), the author differentiates, factorizes dR~/dθ, and shows that, apart from a spurious branch, the maximizer must satisfy a nonlinear equation that is reduced by squaring to a cubic polynomial in cosθmax (Eq. (14)). The paper proves that this cubic has exactly one root in (0,π/4), shows that θmax increases monotonically with the initial speed v0, and derives asymptotic approximations for small and large v0 for both θmax and the maximum range. A separate short-arc argument reproduces the low-speed asymptotics without the cubic.

Significance. If the two gaps identified below are repaired, the paper provides a complete, self-contained analytical solution to a problem previously treated numerically and experimentally. The derivation introduces no fitted parameters; the derivative factorization, the cubic equation, the uniqueness argument, and both asymptotic expansions are correct when checked independently. The asymptotic limits, particularly Rmax ~ v0√((2H+L)/g) for slow starts and Rmax ~ v0²/g + H + (√2−1)L for fast starts, are clean and falsifiable. The paper is a useful reference for instructors and for subsequent work on generalized constrained-motion launch problems. The main novelty is modest—the problem is elementary—but the analysis is rigorous and the presentation is mostly clear.

major comments (2)

[Section III A, paragraph after Eq. (10)] The first factor 2F+cosθ=0 of Eq. (10) is dismissed only for v~0^2 < 1. For 1 ≤ v~0^2 ≤ 5/2 there are admissible physical solutions with cosθ = 2(1−v~0^2)/3, and the paper never rules them out as global maximizers. This matters because the subsequent cubic equation (14) is derived from the second factor of Eq. (10). A one-line argument closes the gap: on this branch cosθ ≤ 0, so in Eq. (9) the terms 2F sinθ cosθ and 2cosθ√(FG) are both nonpositive and R~(θ) ≤ sinθ ≤ 1, while for every v~0^2 ≥ 1 and H~ ≥ 0 the admissible angle θ=π/4 gives R~(π/4) = 1/√2 + F + √(2FG) ≥ √2 > 1. This argument should be inserted after the discussion of Eq. (11).
[Section III A, displayed equation following Eq. (13)] The sign of the (H~+v~0^2)^2 sin^2 θ term is wrong. Substituting F(θ) and G(θ) from Eq. (8) into Eq. (13) yields −(H~+v~0^2)^2 sin^2 θ + (v~0^2−1+cosθ)(H~+v~0^2) − (v~0^2−1+cosθ)^2 cos^2 θ = 0 (equivalently, (H~+v~0^2)^2 cos^2 θ − (H~+v~0^2)^2 + F A − F^2 cos^2 θ = 0). The factorization displayed immediately below, which produces the cubic equation (14), is correct only with this sign corrected.

minor comments (5)

[Section III D, displayed derivative] The derivative of R~(θ) with respect to θ is written as [√2 + 2√H~ α/√(1−α^2)] = 0, but the sign should be minus. With the plus sign the equation has no positive solution and does not lead to αmax = 1/√(2H~+1).
[Equation (20)] The dimensional low-speed maximum range should be Rmax ≃ v0 √{(2H+L)/g}. As printed, '√(2H+L)/g v0' is dimensionally inconsistent unless the square root is understood to extend over (2H+L)/g.
[Section III D] The approximation T~ ≈ 2√H~ for the flight time assumes H~ > 0. For H~ = 0 the maximum is attained at the boundary α = 1, and this case should be mentioned to make the alternative derivation complete.
[Section III A, after Eq. (10)] The phrase 'the angle θ that maximizes R~(θ) satisfies Eq. (10)' is ambiguous; the intended meaning is that it satisfies the second line of Eq. (10), after the first branch has been excluded.
[Typos and phrasing] Please correct the following: 'does not directly existence the unique existence' in Section III A; 'The result indicate' in the caption of Fig. 3; 'eﬀiciency' in the Introduction; and any other typographical errors of this kind.

Circularity Check

0 steps flagged · score 0.0 of 10

No circularity: the optimal-angle cubic and asymptotic results are derived from energy conservation and projectile kinematics with no fitted parameters or self-citation dependency.

full rationale

The paper's central claims are obtained by a self-contained analytical derivation. The range formula in Eq. (4) follows directly from pendulum energy conservation and projectile flight time; nondimensionalization and the auxiliary functions F and G are algebraic redefinitions, not fitted or data-dependent inputs. The derivative of the nondimensional range is computed explicitly, factorized into two branches in Eq. (10), and the physically relevant branch is squared and algebraically reduced to the cubic equation (14). The existence, uniqueness, monotonicity, and asymptotic results are then derived from this cubic using the intermediate value theorem, Descartes' rule of signs, and asymptotic expansions. No parameter is fitted to any subset of data, and no conclusion is assumed as an input. The cited prior work by Bittel, Mungan, Mungan and Lipscombe, and Rave and Sayers is used only for context, comparison of numerical observations, or as a foil for Cardano's formula; none of these citations is load-bearing for the derivation itself. The known incompleteness in excluding the 2F + cos θ = 0 branch for ṽ₀² ≥ 1 is a mathematical correctness gap in the proof of which derivative branch is the global maximizer, not a circularity: the branch is derived from the same exact derivative expression rather than assumed, and the subsequent cubic analysis proceeds from the other factor. A circularity finding would require the derivation to reduce by construction to its own inputs, which is not the case here. The paper is therefore best scored 0 for circularity, with any concerns about rigor assigned to correctness rather than circular dependence.

Assumptions & free parameters 0 free parameters · 4 assumptions · 0 invented entities

The model uses standard Newtonian mechanics and an ideal pendulum. No free parameters are fitted and no new entities are postulated; the quantities L, H, v0, and g are physical inputs of the setup, and the nondimensional parameters are rescalings rather than fitted constants. The central claim rests on ordinary calculus and algebra applied to the equations of motion.

assumptions (4)

domain assumption The pendulum is an ideal simple pendulum with a massless rigid rod, no friction, and no air resistance during the swing.
Invoked in Section II and used to write the conserved speed as v = sqrt(v0^2 - 2gL(1 - cos θ)).
domain assumption After release, the object is a projectile under constant gravitational acceleration g with no air resistance.
Used throughout Section III A to derive the flight time and range in Eqs. (3) and (4).
domain assumption The ground is horizontal and the object lands when its height returns to ground level; the rope and support are ignored after release.
The flight time in Eq. (3) solves the vertical displacement reaching -(H + L(1 - cos θ)).
standard math Standard calculus, the intermediate value theorem, and Descartes' rule of signs are valid.
Used in Section III B to prove existence and uniqueness of the root of f(x) = 0.

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Cite this review

Pith. "Pith review of Theoretical analysis of the maximum range of a projectile released from a pendulum." pith.science (2026). https://pith.science/paper/OM3L2DHE

@misc{pith2026260809157,
  author       = {Pith},
  title        = {Pith review of: Theoretical analysis of the maximum range of a projectile released from a pendulum},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/OM3L2DHE}},
  note         = {Machine review of arXiv:2608.09157}
}

read the original abstract

The motion of a projectile released from a simple pendulum is analyzed, with particular emphasis on investigating the optimal release angle that maximizes the horizontal range and the corresponding maximum range. This system serves as a simplified model of the Tarzan jump problem. Using simple analytical methods, the optimal release angle is shown to be characterized by a cubic equation and to increase with the initial velocity. In addition, asymptotic expressions for both the optimal angle and maximum range are derived in the limits of low and high initial velocity.

Figures

Figures reproduced from arXiv: 2608.09157 by the authors.

**Figure 2.** FIG. 2. Existence of a unique positive solution of [PITH_FULL_IMAGE:figures/full_fig_p005_2.png] view at source ↗

**Figure 3.** (a) shows the relationship between θmax and v˜0 for H˜ = 0, 1, and 2. The values of θmax were obtained numerically by solving the cubic equation f(x) = 0 using Newton’s method with the initial guess x = 1. All numerical calculations in this study were performed using Python. The result indicate that θmax increases with v˜0 for each value of H˜ , with a slower rate of increase for larger H˜ . The dashed lines near v˜… view at source ↗

Figures from the paper (2 more)

**Figure 4.** Figure 4: FIG. 4. Maximum range [PITH_FULL_IMAGE:figures/full_fig_p007_4.png]

**Figure 5.** Figure 5: FIG. 5. Optimal trajectories achieving [PITH_FULL_IMAGE:figures/full_fig_p008_5.png]

Discussion (0). Continue with ORCID to comment.

Reference graph

Works this paper leans on

25 extracted references · 25 canonical work pages

[1]

(11) cannot be realized

Therefore, the solution in Eq. (11) cannot be realized. From the above argument, the angle θ that maximizes ˜R(θ) satisfies Eq. (10). The square root can be eliminated by suitably squaring Eq. (10): F (θ)G(θ) cos2 2θ = (G(θ) − F (θ) cos2 θ)2 sin2 θ. (12) The right-hand side can be further manipulated as (G(θ) − F (θ) cos2 θ)2 sin2 θ = ( ˜H + ˜v2 0 − 2F (θ...

work page
[2]

− (˜v2 0 − 1 + cos θ)2 cos2 θ = 0. This equation still appears complicated, but the left-hand side can be factored relatively easily by treating it as a quadratic in ( ˜H + ˜v2 0): ( ˜H + 1 − cos θ)[cos3 θ + ( ˜H + 2˜v2 0 − 1) cos2 θ − ( ˜H + ˜v2 0)] = 0 . The equation ˜H + 1 − cos θ = 0 is not suitable, since it has no solution with cos θ ≤ 1 when ˜H > 0...

work page
[3]

Equation (14) is the governing equation of θmax and is analyzed in detail in the present study

= 0 , (14) which is a cubic equation in cos θmax. Equation (14) is the governing equation of θmax and is analyzed in detail in the present study. A similar equation was presented by Bittel [7]; however, its derivation process was not provided and its theoretical treatments were not fully explored. One contribution of the present study is the derivation of...

work page
[4]

In 0 < x < 1/ √ 2, no solutions exist because f (x) < 0

and f (1) indicate the existence of at least one solution in 1/ √ 2 < x < 1, whose uniqueness is guaranteed given that f ′(x) > 0 in this interval. In 0 < x < 1/ √ 2, no solutions exist because f (x) < 0. The existence and uniqueness of a positive solution of f (x) = 0 can be shown more concisely using Descartes’ rule of signs [13]. The nonzero coeﬀicient...

work page
[5]

Using x ≃ 1 − c˜v2 0 together with x = cos θmax ≃ 1 − θ2 max/2, the optimal angle is approximated as θmax ≃ √ 2 2 ˜H + 1 ˜v0 = v0√ g(2H + L)

and solving f (x) = 0 up to order ˜v2 0, the coeﬀicient c is found to be c = 1 2 ˜H + 1 . Using x ≃ 1 − c˜v2 0 together with x = cos θmax ≃ 1 − θ2 max/2, the optimal angle is approximated as θmax ≃ √ 2 2 ˜H + 1 ˜v0 = v0√ g(2H + L) . (15) Thus, θmax is proportional to ˜v0 (and hence to v0) in the low-initial-velocity limit. Next, we derive the asymptotic f...

work page
[6]

(7) including a square root

sin θmax = sin θmax[1 − 2 cos θmax + 2 cos2 θmax + 2 ˜H cos θmax + 4˜v2 0 cos θmax], (18) which is much simpler than Eq. (7) including a square root. Note that Eq. (13) holds only when θ = θmax, and thus, this expression for ˜R(θmax) is not correct for other θ ̸= θmax. For ˜v0 ≪ 1, the following approximations are obtained from Eq. (15): sin θmax ≃ θmax ≃...

work page
[7]

Tarzan’s dilemma

yields ˜R(θmax) = 2˜v2 0 + ˜H + √ 2 − 1 + o(˜v0 0), (21) whose dimensional form is R(θmax) ≃ v2 0 g + H + ( √ 2 − 1)L. (22) From Eqs. (20) and (22), the maximum range R(θmax) is proportional to v0 in the regime v0 ≪ 1, whereas it is proportional to v2 0 for v0 ≫ 1. Figure 4 presents numerical results for ˜R(θmax) with ˜H = 0 (circles), 1 (squares), and 2 ...

work page
[8]

yields the same expressions as Eq. (23). The condition ˜v2 0 ≫ ˜H ≫ 1 is equivalent to v2 0/(2g) ≫ H ≫ L. Therefore, the effect of the pendulum, primarily characterized by L, becomes negligible, and the determination of the optimal release angle for Tarzan’s jump becomes indistinguishable from that of a simple projectile launch to order O(2gH/v 2 0). As a...

work page

Show all 25 references

[9]

Ganci and D

S. Ganci and D. Lagomarsino, Eur. J. Phys. 35, 045026 (2014)

2014
[10]

H. Hu, Y. P. Zhao, Y. J. Guo, and M. Y. Zheng, Acta Mech. 223, 441 (2012)

2012
[11]

J. C. Hayen, Int. J. Nonlinear Mech. 38, 357 (2003)

2003
[12]

Cohen, B

C. Cohen, B. Darbois-Texier, G. Dupeux, E. Brunel, D. Quéré, and C. Clanet, Proc. R. Soc. A 470, 20130497 (2014)

2014
[13]

A. K. Lazopoulos and D. Karaoulanis, Axioms 10, 297 (2021)

2021
[14]

K. P. Trout and C. A. Gaston, Phys. Teach. 39, 160 (2001)

2001
[15]

Bittel, Phys

D. Bittel, Phys. Teach. 43, 98 (2005)

2005
[16]

Rave and M

M. Rave and M. Sayers, Phys. Teach. 51, 456 (2013)

2013
[17]

C. E. Mungan and C. Lipscombe, Lat. Am. J. Phys. Educ. 5, 335 (2011)

2011
[18]

Shima, Eur

H. Shima, Eur. J. Phys. 33, 1687 (2012)

2012
[19]

C. E. Mungan, Phys. Teach. 52, 6 (2014)

2014
[20]

Tignol, Galois’ Theory of Algebraic Equations (World Scientific, Singapore, 2001)

J.-P. Tignol, Galois’ Theory of Algebraic Equations (World Scientific, Singapore, 2001)

2001
[21]

Anderson, J

B. Anderson, J. Jackson, and M. Sitharam, Amer. Math. Month. 105, 447 (1998)

1998
[22]

Kovacic and M

I. Kovacic and M. J. Brennan, The Duﬀing Equation (Wiley, New York, 2011)

2011
[23]

Efremidis, K

N. Efremidis, K. Hizanidis, H. E. Nistazakis, D. J. Frantzeskakis, and B. A. Malomed, Phys. Rev. E 62, 7410 (2000)

2000
[24]

A. H. Nayfeh and D. T. Mook, Nonlinear Oscillations (Wiley, New York, 1995)

1995
[25]

Johannessen, Eur

K. Johannessen, Eur. J. Phys. 35, 035014 (2014)

2014

Pith tools

Reviewed August 11, 2026 · model on record in the stance chip above.

[1] [1]

(11) cannot be realized

Therefore, the solution in Eq. (11) cannot be realized. From the above argument, the angle θ that maximizes ˜R(θ) satisfies Eq. (10). The square root can be eliminated by suitably squaring Eq. (10): F (θ)G(θ) cos2 2θ = (G(θ) − F (θ) cos2 θ)2 sin2 θ. (12) The right-hand side can be further manipulated as (G(θ) − F (θ) cos2 θ)2 sin2 θ = ( ˜H + ˜v2 0 − 2F (θ...

work page

[2] [2]

− (˜v2 0 − 1 + cos θ)2 cos2 θ = 0. This equation still appears complicated, but the left-hand side can be factored relatively easily by treating it as a quadratic in ( ˜H + ˜v2 0): ( ˜H + 1 − cos θ)[cos3 θ + ( ˜H + 2˜v2 0 − 1) cos2 θ − ( ˜H + ˜v2 0)] = 0 . The equation ˜H + 1 − cos θ = 0 is not suitable, since it has no solution with cos θ ≤ 1 when ˜H > 0...

work page

[3] [3]

Equation (14) is the governing equation of θmax and is analyzed in detail in the present study

= 0 , (14) which is a cubic equation in cos θmax. Equation (14) is the governing equation of θmax and is analyzed in detail in the present study. A similar equation was presented by Bittel [7]; however, its derivation process was not provided and its theoretical treatments were not fully explored. One contribution of the present study is the derivation of...

work page

[4] [4]

In 0 < x < 1/ √ 2, no solutions exist because f (x) < 0

and f (1) indicate the existence of at least one solution in 1/ √ 2 < x < 1, whose uniqueness is guaranteed given that f ′(x) > 0 in this interval. In 0 < x < 1/ √ 2, no solutions exist because f (x) < 0. The existence and uniqueness of a positive solution of f (x) = 0 can be shown more concisely using Descartes’ rule of signs [13]. The nonzero coeﬀicient...

work page

[5] [5]

Using x ≃ 1 − c˜v2 0 together with x = cos θmax ≃ 1 − θ2 max/2, the optimal angle is approximated as θmax ≃ √ 2 2 ˜H + 1 ˜v0 = v0√ g(2H + L)

and solving f (x) = 0 up to order ˜v2 0, the coeﬀicient c is found to be c = 1 2 ˜H + 1 . Using x ≃ 1 − c˜v2 0 together with x = cos θmax ≃ 1 − θ2 max/2, the optimal angle is approximated as θmax ≃ √ 2 2 ˜H + 1 ˜v0 = v0√ g(2H + L) . (15) Thus, θmax is proportional to ˜v0 (and hence to v0) in the low-initial-velocity limit. Next, we derive the asymptotic f...

work page

[6] [6]

(7) including a square root

sin θmax = sin θmax[1 − 2 cos θmax + 2 cos2 θmax + 2 ˜H cos θmax + 4˜v2 0 cos θmax], (18) which is much simpler than Eq. (7) including a square root. Note that Eq. (13) holds only when θ = θmax, and thus, this expression for ˜R(θmax) is not correct for other θ ̸= θmax. For ˜v0 ≪ 1, the following approximations are obtained from Eq. (15): sin θmax ≃ θmax ≃...

work page

[7] [7]

Tarzan’s dilemma

yields ˜R(θmax) = 2˜v2 0 + ˜H + √ 2 − 1 + o(˜v0 0), (21) whose dimensional form is R(θmax) ≃ v2 0 g + H + ( √ 2 − 1)L. (22) From Eqs. (20) and (22), the maximum range R(θmax) is proportional to v0 in the regime v0 ≪ 1, whereas it is proportional to v2 0 for v0 ≫ 1. Figure 4 presents numerical results for ˜R(θmax) with ˜H = 0 (circles), 1 (squares), and 2 ...

work page

[8] [8]

yields the same expressions as Eq. (23). The condition ˜v2 0 ≫ ˜H ≫ 1 is equivalent to v2 0/(2g) ≫ H ≫ L. Therefore, the effect of the pendulum, primarily characterized by L, becomes negligible, and the determination of the optimal release angle for Tarzan’s jump becomes indistinguishable from that of a simple projectile launch to order O(2gH/v 2 0). As a...

work page