Fractional Differential Equations in Sports Training in Universities

If the air resistance is negligible, we will throw the object obliquely upward at a certain rate. What is the farthest distance when the elevation angle is? The answer in middle school physics is 45°. The situation is different when putting shot puts. The throwing point of the shot is not on the ground but at a height above the ground [1]. In the sports-related instruction book, only the best projectile angle is 38°–42°, so it isn't easy to guide sports training correctly. Students cannot integrate theoretical models with actual sports. According to the projectile theory, this paper makes a theoretical calculation and analysis on the shot's movement [2]. At the same time, experimental simulation verification and discussion were carried out.

Approximate solutions in differential equations and parametric equations for calculating the best throwing angle [3]. From Figure 1, we can see that if υ_y = υ_x tan θ, assumes tan θ = q, then there is υ_y = qυ_x, dυ_y = qdυ_x + υ_xdq. (1) dυydυx=υyυx+mgkυxυx2+υy2 {{d{\upsilon _y}} \over {d{\upsilon _x}}} = {{{\upsilon _y}} \over {{\upsilon _x}}} + {{mg} \over {k{\upsilon _x}\sqrt {\upsilon _x^2 + \upsilon _y^2} }} We can get 1+q2 dq= mgkυx3dυx \sqrt {1 + {q^2}} \,dq = \,{{mg} \over {k\upsilon _x^3}}d{\upsilon _x} by substituting υ_y = qυ_x, dυ_y = qdυ_x + υ_xdq into the above formula and simplifying it.

Figure 1

The integral on both sides can be obtained ∫q0q1+q2dq=∫υx0υxmgkυx3dυx \int_{{q_0}}^q {\sqrt {1 + {q^2}} dq = \int_{{\upsilon _{{x_0}}}}^{{\upsilon _x}} {{{mg} \over {k\upsilon _x^3}}d{\upsilon _x}} }

So there is (2) 12[q1+q2+ln(q+1+q2)]−12[q01+q2+ln(q0+1+q2)]=mg2k(1υx02−1υx2) \matrix{ {{1 \over 2}\left[ {q\sqrt {1 + {q^2}} + \ln \left( {q + \sqrt {1 + {q^2}} } \right)} \right] - } \hfill \cr {{1 \over 2}\left[ {{q_0}\sqrt {1 + {q^2}} + \ln \left( {{q_0} + \sqrt {1 + {q^2}} } \right)} \right] = } \hfill \cr {{{mg} \over {2k}}\left( {{1 \over {\upsilon _{{x_0}}^2}} - {1 \over {\upsilon _x^2}}} \right)} \hfill \cr } q₀ = tan θ₀ = tan α, υ_x0 = υ₀ cos α in the above formula. Where α and υ₀ are the angle and initial velocity of the throw, respectively. (3) f(q)=q1+q2+ln(q+1+q2) {f_{\left( q \right)}} = q\sqrt {1 + {q^2}} + \ln \left( {q + \sqrt {1 + {q^2}} } \right)

Equation (2) can be abbreviated as (4) 1υx02−1υx2=kmg[f(q)−f(q0)] {1 \over {\upsilon _{{x_0}}^2}} - {1 \over {\upsilon _x^2}} = {k \over {mg}}\left[ {{f_{\left( q \right)}} - {f_{\left( {{q_0}} \right)}}} \right]

From equation (4) and υ_y = qυ_x, we can get (5) υx=1C−kf(q)mg=1C(1−kf(q)mgC) {\upsilon _x} = {1 \over {\sqrt {C - {{k{f_{\left( q \right)}}} \over {mg}}} }} = {1 \over {\sqrt {C\left( {1 - {{k{f_{\left( q \right)}}} \over {mg{\rm{C}}}}} \right)} }} (6) υy=qC−kf(q)mg=qC(1−kf(q)mgC) {\upsilon _y} = {q \over {\sqrt {C - {{k{f_{\left( q \right)}}} \over {mg}}} }} = {q \over {\sqrt {C\left( {1 - {{k{f_{\left( q \right)}}} \over {mg{\rm{C}}}}} \right)} }}

In the formula, the constant C=1υx02+kf(q)mg=1υx02+kmg[q01+q02+ln(q0+1+q02)] C = {1 \over {\upsilon _{{x_0}}^2}} + {{k{f_{\left( q \right)}}} \over {mg}} = {1 \over {\upsilon _{{x_0}}^2}} + {k \over {mg}}\left[ {{q_0}\sqrt {1 + q_0^2} + \ln \left( {{q_0} + \sqrt {1 + q_0^2} } \right)} \right] is determined by υ₀, α, k, m. It can be seen from equations (5) and (6) that it is difficult to obtain the relationship between υ_x and υ_y without approximation, so it is impossible to solve υ_x and υ_y from equations (1) and (2). The constant C in (5) and (6) can be written as C=1υx02+kf(q0)mg=1υx02+(1+kυx02f(q0)mg) C = {1 \over {\upsilon _{{x_0}}^2}} + {{k{f_{\left( {{q_0}} \right)}}} \over {mg}} = {1 \over {\upsilon _{{x_0}}^2}} + \left( {1 + {{k\upsilon _{{x_0}}^2{f_{\left( {{q_0}} \right)}}} \over {mg}}} \right) , so kf(q0)mgC=kυx02f(q0)mg(1+kυx02f(q0)mg) {{k{f_{\left( {{q_0}} \right)}}} \over {mgC}} = {{k\upsilon _{{x_0}}^2{f_{\left( {{q_0}} \right)}}} \over {mg\left( {1 + {{k\upsilon _{{x_0}}^2{f_{\left( {{q_0}} \right)}}} \over {mg}}} \right)}} . In this way, we can prove that this item is a small amount under certain conditions.

In general, it can be assumed that the air resistance at the initial moment is a small amount compared to gravity [4]. That is, there is kυx02mg1 {{k\upsilon _{{x_0}}^2} \over {mg}}1 . And because the throwing angle α is generally less than 45° when the hand is shot, f(q0)<2.3,1+kυx02f(q0)mg≈1+2.3kυx02mg≈1 {f_{\left( {{q_0}} \right)}} < 2.3,1 + {{k\upsilon _{{x_0}}^2{f_{\left( {{q_0}} \right)}}} \over {mg}} \approx 1 + {{2.3k\upsilon _{{x_0}}^2} \over {mg}} \approx 1 can be obtained by f_(q0) substituting q₀ = tan α < 1 into the expression.

In the rising phase, θ is always less than α, so there is q < q₀. Since f_(q) is an increasing function of E when q > 0, there is f_(q) < f_(q0) < 2.3.

In the descending stage, |θ| = 70°, |f_(q)| is negative, but even |θ| = 70°, |f_(q)| is only 9.8. Therefore, kυx02f(q)mg {{k\upsilon _{{x_0}}^2{f_{\left( q \right)}}} \over {mg}} can still be considered to be much smaller than 1. The above analysis shows that in general, kf(q)mgC≈kυx02f(q)mg {{k{f_{\left( q \right)}}} \over {mgC}} \approx {{k\upsilon _{{x_0}}^2{f_{\left( q \right)}}} \over {mg}} can be regarded as a small amount [5]. We can use the series expansion formula (1 + x)^m ≈ 1 + mx to obtain (7) υx≈1C(1+k2mgCf(q)) {\upsilon _x} \approx {1 \over {\sqrt C }}\left( {1 + {k \over {2mgC}}{f_{\left( q \right)}}} \right) (8) υy≈qC(1+k2mgCf(q)) {\upsilon _y} \approx {q \over {\sqrt C }}\left( {1 + {k \over {2mgC}}{f_{\left( q \right)}}} \right)

Although the above is a computational approximation, it is impossible to obtain a direct relationship between υ_x and υ_y due to function complexity f_(q). The relationship between q and υ_x or υ_y cannot be expressed by elementary functions, so it is impossible to Substitute formula (1) and formula (2) to find the relationship between υ_x and υ_y over time. But if we can find the relationship between q and t. In this way, the solutions of υ_x and υ_y are found at least in the form [6]. We substitute υ_y = qυ_x into the left side of equation (1) and take dυxdt=dυxdqdqdt {{d{\upsilon _x}} \over {dt}} = {{d{\upsilon _x}} \over {dq}}{{dq} \over {dt}} into account. Then dυxdq=k1+q2mgCC(1−kf(q)mgC)−32 {{d{\upsilon _x}} \over {dq}} = {{k\sqrt {1 + {q^2}} } \over {mgC\sqrt C }}{\left( {1 - {{k{f_{\left( q \right)}}} \over {mgC}}} \right)^{ - {3 \over 2}}} can be obtained by formula (5).

υ_x does not use formula (7) to be more precise. In addition, the derivative f(q)′=21+q2 f_{\left( q \right)}^\prime = 2\sqrt {1 + {q^2}} of the function f_(q) is substituted into formula (1) to obtain −kυx21+q2=mk1+q2mgCC(1−kf(q)mgC)−32dqdt - k\upsilon _x^2\sqrt {1 + {q^2}} = m{{k\sqrt {1 + {q^2}} } \over {mgC\sqrt C }}{\left( {1 - {{k{f_{\left( q \right)}}} \over {mgC}}} \right)^{ - {3 \over 2}}}{{dq} \over {dt}} . From equation (5), we can see 1υx2=C(1−kf(q)mgC) {1 \over {\upsilon _x^2}} = C\left( {1 - {{k{f_{\left( q \right)}}} \over {mgC}}} \right) . Substituting into the above formula and using series expansion, we can get (9) −gCdt=(1−kf(q)mgC)−12dq≈(1+kf(q)2mgC)dq - g\sqrt C dt = {\left( {1 - {{k{f_{\left( q \right)}}} \over {mgC}}} \right)^{ - {1 \over 2}}}dq \approx \left( {1 + {{k{f_{\left( q \right)}}} \over {2mgC}}} \right)dq

The integral on both sides can be obtained (10) −gCt=q−q0+k2mgC∫q0qf(q)dq - g\sqrt C t = q - {q_0} + {k \over {2mgC}}\int_{{q_0}}^q {{f_{\left( q \right)}}dq}

Function f_(q) consists of two parts: f1(q)=q1+q2 {f_{1\left( q \right)}} = q\sqrt {1 + {q^2}} , f2(q)=ln(q+1+q2) {f_{2\left( q \right)}} = \ln \left( {q + \sqrt {1 + {q^2}} } \right) . In the formula ln(q+1+q2)= arsinhq \ln \left( {q + \sqrt {1 + {q^2}} } \right) = \,ar\sinh \,q . The following integral will use this equation. ∫q0qf(q)dq=∫q0qf1(q)dq+∫q0qf2(q)dq=∫q0qq1+q2dq+∫q0qar sinh qdq=[13(1+q2)32+qar sinh q−1+q2]−[13(1+q02)32+q0 ar sinh q0−1+q2]=φ(q)−φ(q0) \matrix{ {\int_{{q_0}}^q {{f_{\left( q \right)}}dq = } \int_{{q_0}}^q {{f_{1\left( q \right)}}dq + \int_{{q_0}}^q {{f_{2\left( q \right)}}dq = } } } \hfill \cr {\int_{{q_0}}^q {q\sqrt {1 + {q^2}} dq + \int_{{q_0}}^q {ar\,\sinh \,qdq = } } } \hfill \cr {\left[ {{1 \over 3}{{\left( {1 + {q^2}} \right)}^{{3 \over 2}}} + qar\,\sinh \,q - \sqrt {1 + {q^2}} } \right] - } \hfill \cr {\left[ {{1 \over 3}{{\left( {1 + q_0^2} \right)}^{{3 \over 2}}} + {q_0}\,ar\,\sinh \,{q_0} - \sqrt {1 + {q^2}} } \right] = {\varphi _{\left( q \right)}} - {\varphi _{\left( {{q_0}} \right)}}} \hfill \cr } can be obtained from the nature of the integral. Where φ(q)=[13(1+q02)32+qar sinh q−1+q2] {\varphi _{\left( q \right)}} = \left[ {{1 \over 3}{{\left( {1 + q_0^2} \right)}^{{3 \over 2}}} + qar\,\sinh \,q - \sqrt {1 + {q^2}} } \right] . So from equation (10) we can get (11) t=q0−qgC+k2mg2CC[φ(q0)−φ(q)] t = {{{q_0} - q} \over {g\sqrt C }} + {k \over {2m{g^2}C\sqrt C }}\left[ {{\varphi _{\left( {{q_0}} \right)}} - {\varphi _{\left( q \right)}}} \right]

Formally, the relationship between q and time can be obtained from equation (11) q = ψ(t). In principle, the relationship between sub-velocity and time has been derived from equations (5), (6), and (11). The coordinates x and y are calculated below the relationship with q or t. dxdt=υx=1C(1−kmgCf(q))−12 {{dx} \over {dt}} = {\upsilon _x} = {1 \over {\sqrt C }}{\left( {1 - {k \over {mgC}}{f_{\left( q \right)}}} \right)^{ - {1 \over 2}}} can be obtained by formula (5). So dx=1C(1−kmgCf(q))−12dt dx = {1 \over {\sqrt C }}{\left( {1 - {k \over {mgC}}{f_{\left( q \right)}}} \right)^{ - {1 \over 2}}}dt . From formula (8), dt=−1gC(1−kmgCf(q))−12dq dt = {{ - 1} \over {g\sqrt C }}{\left( {1 - {k \over {mgC}}{f_{\left( q \right)}}} \right)^{ - {1 \over 2}}}dq can be obtained. Substitute the above formula to get dx=−1gC(1−kmgCf(q))−1dq dx = {{ - 1} \over {gC}}{\left( {1 - {k \over {mgC}}{f_{\left( q \right)}}} \right)^{ - 1}}dq . It can be obtained by series expansion. dx≈−1gC(1−kmgCf(q))dq dx \approx {{ - 1} \over {gC}}\left( {1 - {k \over {mgC}}{f_{\left( q \right)}}} \right)dq

Integrate both sides to get x=1gC[(q0−q)+kmgC(φ(q0)−φ(q))] x = {1 \over {gC}}\left[ {\left( {{q_0} - q} \right) + {k \over {mgC}}\left( {{\varphi _{\left( {{q_0}} \right)}} - {\varphi _{\left( q \right)}}} \right)} \right] . From equation (9), kmgC(φ(q0)−φ(q))=2[gtC−(q0−q)] {k \over {mgC}}\left( {{\varphi _{\left( {{q_0}} \right)}} - {\varphi _{\left( q \right)}}} \right) = 2\left[ {gt\sqrt C - \left( {{q_0} - q} \right)} \right] can be obtained. Substitute the above formula to get (12) x=1gC[(q0−q)+kmgC(φ(q0)−φ(q))]=1gC[2gtC−(q0−q)] x = {1 \over {gC}}\left[ {\left( {{q_0} - q} \right) + {k \over {mgC}}\left( {{\varphi _{\left( {{q_0}} \right)}} - {\varphi _{\left( q \right)}}} \right)} \right] = {1 \over {gC}}\left[ {2gt\sqrt C - \left( {{q_0} - q} \right)} \right] When k = 0, q0−q=gtC {q_0} - q = gt\sqrt C can be known from equation (11). Therefore, x=tC=υx0t=υ0t cos α x = {t \over {\sqrt C }} = {\upsilon _{{x_0}}}t = {\upsilon _0}t\,\cos \,\alpha can be obtained from equation (12). This is completely consistent with the conclusion when there is no air resistance. Because dυydυx=q {{d{\upsilon _y}} \over {d{\upsilon _x}}} = q , there must be dydx=dυydυx=q {{dy} \over {dx}} = {{d{\upsilon _y}} \over {d{\upsilon _x}}} = q , that is, dy = qdx.

We can get dy=−qgC(1−kmgCf(q))−1 dq≈−qgC(1+kmgCf(q))dq dy = {{ - q} \over {gC}}{\left( {1 - {k \over {mgC}}{f_{\left( q \right)}}} \right)^{ - 1}}\,dq \approx {{ - q} \over {gC}}\left( {1 + {k \over {mgC}}{f_{\left( q \right)}}} \right)dq by substituting dx=−1gC(1−kmgCf(q))−1 dq dx = {{ - 1} \over {gC}}{\left( {1 - {k \over {mgC}}{f_{\left( q \right)}}} \right)^{ - 1}}\,dq into dy = qdx. The integral on both sides can be obtained (13) y=1gC[q02−q22+kmgC(λ(q0)−λ(q)] y = {1 \over {gC}}\left[ {{{q_0^2 - {q^2}} \over 2} + {k \over {mgC}}\left( {{\lambda _{\left( {{q_0}} \right)}} - {\lambda _{\left( q \right)}}} \right.} \right]

Among them, λ_(q) is a function introduced for the convenience of writing [7]. It appears in the second part of the integral ∫q0qqf(q)dq \int_{{q_0}}^q {q{f_{\left( q \right)}}dq} on the right side of the dy equation. d[q2f(q)]=2qf(q)dq+q2f(q)′ dq=2qf(q)dq+2q21+q2dq d\left[ {{q^2}{f_{\left( q \right)}}} \right] = 2q{f_{\left( q \right)}}dq + {q^2}f_{\left( q \right)}^\prime\,dq = 2q{f_{\left( q \right)}}dq + 2{q^2}\sqrt {1 + {q^2}} dq can be known from the differential knowledge.

So ∫q0qqf(q)dq=12[q2f(q)−q02f(q0)]−∫q0qq21+q2dq=12[q2f(q)−q02f(q0)]−14[q(1+q2)32−q0(1+q02)32]+14∫q0q1+q2dq \matrix{ {\int_{{q_0}}^q {q{f_{\left( q \right)}}dq = {1 \over 2}\left[ {{q^2}{f_{\left( q \right)}} - q_0^2{f_{\left( {{q_0}} \right)}}} \right] - } \int_{{q_0}}^q {{q^2}\sqrt {1 + {q^2}} dq = } } \hfill \cr {{1 \over 2}\left[ {{q^2}{f_{\left( q \right)}} - q_0^2{f_{\left( {{q_0}} \right)}}} \right] - {1 \over 4}\left[ {q{{\left( {1 + {q^2}} \right)}^{{3 \over 2}}} - } \right.} \hfill \cr {\left. {{q_0}{{\left( {1 + q_0^2} \right)}^{{3 \over 2}}}} \right] + {1 \over 4}\int_{{q_0}}^q {\sqrt {1 + {q^2}} dq} } \hfill \cr } . From the formula (5) and the definition of f_(q), ∫q0q1+q2dq=12(f(q)−f(q0)) \int_{{q_0}}^q {\sqrt {1 + {q^2}} dq = {1 \over 2}\left( {{f_{\left( q \right)}} - {f_{\left( {{q_0}} \right)}}} \right)} can be known.

Substituting into the above formula and simplification appropriately ∫q0qqf(q)dq=12[(q2+14)f(q)−12q(1+q2)32]−12[(q02+14)f(q0)−12q0(1+q02)32]=λ(q)−λ(q0) \matrix{ {\int_{{q_0}}^q {q{f_{\left( q \right)}}dq = {1 \over 2}\left[ {\left( {{q^2} + {1 \over 4}} \right){f_{\left( q \right)}} - {1 \over 2}q{{\left( {1 + {q^2}} \right)}^{{3 \over 2}}}} \right] - } } \hfill \cr {{1 \over 2}\left[ {\left( {q_0^2 + {1 \over 4}} \right){f_{\left( {{q_0}} \right)}} - {1 \over 2}{q_0}{{\left( {1 + q_0^2} \right)}^{{3 \over 2}}}} \right] = {\lambda _{\left( q \right)}} - {\lambda _{\left( {{q_0}} \right)}}} \hfill \cr } can be obtained. Where λ(q)=12[(q02+14)f(q)−12q(1+q2)32] {\lambda _{\left( q \right)}} = {1 \over 2}\left[ {\left( {q_0^2 + {1 \over 4}} \right){f_{\left( q \right)}} - {1 \over 2}q{{\left( {1 + {q^2}} \right)}^{{3 \over 2}}}} \right] . When k = 0, y=q02−q22gC y = {{q_0^2 - {q^2}} \over {2gC}} can be obtained by formula (13). q=q0−gtC q = {q_0} - gt\sqrt C can be obtained from formula (11). Substitute and note C=1υx02,q0υx0=υy0=υ0sinα C = {1 \over {\upsilon _{{x_0}}^2}},{q_0}{\upsilon _{{x_0}}} = {\upsilon _{{y_0}}} = {\upsilon _0}\sin \alpha . You can get y=υ0t sin a−12gt2 y = {\upsilon _0}t\,\sin \,a - {1 \over 2}g{t^2} . This is consistent with the formula when there is no resistance.

So far, we have derived the exact expressions (5) and (6) of υ_x and υ_y and q. At the same time, we find the approximate expressions (11), (12), (13) of t, x, y and q. However, since the parameter q is difficult to eliminate, it is difficult to derive the direct relationship between υ_x, υ_y, x, y and t. So this can only be said to be obtained. The formal solution of the differential equation can be obtained instead of the real solution. Because of this, it is impossible to derive the equation of the best throwing angle. However, computer numerical calculations can be used when the basic parameter α υ₀, k, m is known. Given any time t, q can be calculated from equation (6) and then substituted into equations (5), (6), (12), (13) to calculate υ_x, υ_y, x, y.

Assuming that the horizontal distance between the landing point and the throwing point is s, the value of q when landing is represented by Q. Then h = h₁ + L sin a can be seen in Figure 1. Among them, h1 is the height of the shoulder, L is the arm length. The y axis coordinate when landing is y = − h. Assume Q₀ = q₀ = tan a. From equations (12) and (13), we can get (14) s=1gC[(Q0−Q)+kmgC(φ(Q0)−φ(Q))] s = {1 \over {gC}}\left[ {\left( {{Q_0} - Q} \right) + {k \over {mgC}}\left( {{\varphi _{\left( {{Q_0}} \right)}} - {\varphi _{\left( Q \right)}}} \right)} \right] (15) −(h1+L sin a)=1gC[Q02−Q22+kmgC(λ(Q0)−λ(Q)] - \left( {{h_1} + L\,\sin \,a} \right) = {1 \over {gC}}\left[ {{{Q_0^2 - {Q^2}} \over 2} + {k \over {mgC}}\left( {{\lambda _{\left( {{Q_0}} \right)}} - {\lambda _{\left( Q \right)}}} \right.} \right]

When the basic parameter h₁, L, υ₀, m is given, it can be seen from equation (15) that Q is uniquely determined by a. At this time, given a value of a, the computer can calculate a value of Q. The value of D can be calculated by substituting equation (14). If the value range of a is 36°–45°, these 10-degree values (36°, 37°,…, 45°) can be substituted into equation (15). Calculate 10 Q values. Substitute equation (15) to calculate 10 s values. Among them, the largest s value (represented by sm) and its corresponding a value (represented by am) are the best throwing angles [8]. Therefore, equations (14) and (15) are the best throwing angles we want to derive from calculating the oblique upward parabola. The most important parameter equation for the best throw angle.

This is basically in line with the experimental results. At the same time, the experiment also proves that the results of theoretical calculation and analysis have a certain guiding effect on the shot put. The coach can guide the athletes to master the best shot angle based on the theoretical calculation results and by measuring the athlete's shot height and initial speed.