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Fractional Differential Equations in Sports Training in Universities

Online veröffentlicht: 15 Jul 2022
Volumen & Heft: AHEAD OF PRINT
Seitenbereich: -
Eingereicht: 16 Jan 2022
Akzeptiert: 24 Mar 2022
Zeitschriftendaten
License
Format
Zeitschrift
eISSN
2444-8656
Erstveröffentlichung
01 Jan 2016
Erscheinungsweise
2 Hefte pro Jahr
Sprachen
Englisch
Introduction

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.

The parametric equation and practical equation of the best throwing angle of shot put

Approximate solutions in differential equations and parametric equations for calculating the best throwing angle [3]. From Figure 1, we can see that if υy = υx tan θ, assumes tan θ = q, then there is υy = x, y = qdυx + υxdq. 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+q2dq=mgkυx3dυx \sqrt {1 + {q^2}} \,dq = \,{{mg} \over {k\upsilon _x^3}}d{\upsilon _x} by substituting υy = x, y = qdυx + υxdq into the above formula and simplifying it.

Figure 1

Angle relationship when the shot is thrown

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 12[q1+q2+ln(q+1+q2)]12[q01+q2+ln(q0+1+q2)]=mg2k(1υx021υx2) \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, υx0 = υ0 cos α in the above formula. Where α and υ0 are the angle and initial velocity of the throw, respectively. 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 1υx021υx2=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 υy = x, we can get υx=1Ckf(q)mg=1C(1kf(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)} }} υy=qCkf(q)mg=qC(1kf(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 υx and υy without approximation, so it is impossible to solve υx and υy 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)mg1+2.3kυx02mg1 {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)mgCkυx02f(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 υx1C(1+k2mgCf(q)) {\upsilon _x} \approx {1 \over {\sqrt C }}\left( {1 + {k \over {2mgC}}{f_{\left( q \right)}}} \right) υyqC(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 υx and υy due to function complexity f(q). The relationship between q and υx or υy cannot be expressed by elementary functions, so it is impossible to Substitute formula (1) and formula (2) to find the relationship between υx and υy over time. But if we can find the relationship between q and t. In this way, the solutions of υx and υy are found at least in the form [6]. We substitute υy = 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(1kf(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).

υx 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(1kf(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(1kf(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 gCdt=(1kf(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 gCt=qq0+k2mgCq0qf(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+q0qarsinhqdq=[13(1+q2)32+qarsinhq1+q2][13(1+q02)32+q0arsinhq01+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+qarsinhq1+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 t=q0qgC+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(1kmgCf(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(1kmgCf(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(1kmgCf(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(1kmgCf(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. dx1gC(1kmgCf(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[(q0q)+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(q0q)] {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 x=1gC[(q0q)+kmgC(φ(q0)φ(q))]=1gC[2gtC(q0q)] 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, q0q=gtC {q_0} - q = gt\sqrt C can be known from equation (11). Therefore, x=tC=υx0t=υ0tcosα 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(1kmgCf(q))1dqqgC(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(1kmgCf(q))1dq 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 y=1gC[q02q22+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)32q0(1+q02)32]+14q0q1+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=q02q22gC y = {{q_0^2 - {q^2}} \over {2gC}} can be obtained by formula (13). q=q0gtC 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=υ0tsina12gt2 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 υx and υy 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 υx, υy, 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 υx, υy, 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 s=1gC[(Q0Q)+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] (h1+Lsina)=1gC[Q02Q22+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.

Experimental Design
Experimental design ideas

High-speed photography is often used to record movements in sports biomechanics research. At the same time, we use instruments to measure the relationship between force, speed, acceleration, and joint angle changes [9]. It is processed by force analysis, computer processing, and other methods. In general, high-precision instruments cannot be used for measurement. For this reason, we designed a simulation experiment: the use of fluid jets to simulate the projectile's movement to simulate and fit the actual movement of the shot to reflect the real movement of the projectile indirectly. This verifies the feasibility of theoretical calculation and analysis.

Experimental simulation of the projectile motion of the shot put requires the design of experiments by determining some parameters such as the object's initial velocity and initial height. Speed control is shown in Figure 2.

Figure 2

Speed control simulation

The height of the selected water outlet is also zero. The liquid surface velocity is zero. The height difference between the water outlet and the liquid surface is h'. The air pressure between the water outlet and the liquid surface is atmospheric p0. It is obtained from the above formula ρgh+p0=12ρυ02+p0 \rho gh' + {p_0} = {1 \over 2}\rho \upsilon _0^2 + {p_0}

In the formula υ0 is the water outlet flow rate (ie, the initial velocity of the projectile) and ρ is the liquid density. Solve for υ0=2gh {\upsilon _0} = \sqrt {2gh'} from this. As the jet liquid level drops, the flow velocity at the outlet will gradually decrease. The liquid ejected from the water outlet does not flow steadily, but considering that the free surface of the liquid has a large diameter, the water outlet has a very small diameter [10]. If the observation time is short, there is no significant change in the liquid level, and it can still be regarded as a steady flow. Considering the steady flow speed and air resistance, we use a ballpoint pen jacket to insert one end of an elastic rubber tube as a water outlet to make an easy jet.

Equipment and device installation

Table (0.8m high), 2 elevated platforms (one is about 1m high, and the other is about 0.8m high). Elastic rubber tube (about 2m in length), clean ballpoint pen jacket (about 0.3cm in diameter). Protractor (inner radius 5), meter ruler, bucket, and clean carpet.

1) We insert the ballpoint pen jacket into one end of the elastic rubber tube as the water outlet to make an easy jet. 2) Place the bucket on the elevated platform and fill it with water until the water surface is flush with the upper port diameter of the bucket. 3) Insert one end (water inlet) of the non-ballpoint pen jacket into the bucket. The water outlet sags to a certain height (here, 0.8m is chosen) to be close to the table's edge as the water outlet point.

Experimental procedure

1) Insert the water inlet into the bucket, and let the water outlet fall freely. When the water flow is normal, plug the outlet end and prepare for the experiment. 2) Use a protractor to control the ballpoint pen jacket to select the water outlet angle. It shoots out from a height of 0.8m. 3) To prevent inaccurate positioning of water splash, we choose a clean carpet to spread on the experimental site. Use small objects to mark where the waterfalls and record them. 4) Use a meter to measure the horizontal run length of the water flow. Adjust the height of the liquid level for the next round of experiments.

Experimental data

The experimental data are shown in Table 1 and Table 2.

Test data 1

Experiment number Height from liquid level to outlet h′/m Ejection height h/m Exit velocity v0/(m·s−1)
Experiment 1 1.473 0.8 5.373
Experiment 2 1.008 0.8 4.445

Test data 2

Experiment 1 Exit angle α/(°) 25 35 38 40 45 50
Outgoing distance s/m 2.324 2.437 2.5 2.36 2.726 2.23
Experiment 2 Exit angle α/(°) 20 30 35 40 45
Outgoing distance s/m 1.77 1.815 1.82 1.754 1.73

It can be seen from Table 2 that the best spray angles of experiment 1 and experiment 2 are about 38° and 35°, respectively. According to the theoretical analysis formula αopt=arcsinυ022gh+2υ02 {\alpha _{opt}} = \arcsin \sqrt {{{\upsilon _0^2} \over {2gh + 2\upsilon _0^2}}} , αopt1 = 38.83° in experiment 1 can be calculated, and αopt2=36.75° in experiment 2. This is in good agreement with the experimental results, which proves the correctness of the optimal jet angle formula.

Conclusion

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.

Figure 1

Angle relationship when the shot is thrown
Angle relationship when the shot is thrown

Figure 2

Speed control simulation
Speed control simulation

Test data 2

Experiment 1 Exit angle α/(°) 25 35 38 40 45 50
Outgoing distance s/m 2.324 2.437 2.5 2.36 2.726 2.23
Experiment 2 Exit angle α/(°) 20 30 35 40 45
Outgoing distance s/m 1.77 1.815 1.82 1.754 1.73

Test data 1

Experiment number Height from liquid level to outlet h′/m Ejection height h/m Exit velocity v0/(m·s−1)
Experiment 1 1.473 0.8 5.373
Experiment 2 1.008 0.8 4.445

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