In the Olympiad of 708 B.C., some athletes competing in the standing long jump used handheld weights called halteres to lengthen their jumps (Fig. 9-56).The weights were swung up in front just before liftoff and then swung down and thrown backward during the flight. Suppose a modern 78 kg long jumper similarly uses two 5.50 kg halters, throwing them horizontally to the rear at his maximum height such that their horizontal velocity is zero relative to the ground. Let his liftoff velocity be $\overrightarrow{\mathbf{v}}\mathbf{=}\mathbf{(}\mathbf{9}\mathbf{.}\mathbf{5}\hat{\mathbf{i}}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{)}$with or without the halters, and assume that he lands at the liftoff level. What distance would the use of the halters add to his range?

The lift off velocity is,${\overrightarrow{\mathrm{v}}}_0=9.5\hat{\mathrm{i}}+4.0\hat{\mathrm{j}}\mathrm{m}/\mathrm{s}$

The two halters of mass m = 5.50 kg .

Jumper’s mass is M = 78 kg .

By using the given initial lift off speed, find the rangeof the athlete without using halters. By applying conservation of energy, find the distance that the use of the halters would add to his range. According to conservation of energy, energy can neither be created, nor be destroyed.

Formulae are as follow:

The range of the athlete without using halters is,$R_0=\frac{v_0^2\sin 2{\theta }_0}{g}............\left(1\right)$

The conservation of momentum is,$\left(M+2m\right)v_{x0}=Mv{\text{'}}_x.......\left(2\right)$

The increase in the range is,$∆R=\left(∆v{\text{'}}_x\right)t.......\left(3\right)$

With ${\overrightarrow{v}}_0=9.5\hat{i}+4.0\hat{j}\mathrm{m}/\mathrm{s}$, the initial speed is,

$$\begin{gathered} {\overrightarrow{v}}_0=\sqrt{{v_{x0}}^2+{v_{y0}}^2} \\ {v}_0=\sqrt{9.5^2+4.0^2} \\ {v}_0=10.31\mathrm{m}/\mathrm{s} \end{gathered}$$

And take off angle of the athlete is,

$$\begin{gathered} {\theta }_0={\tan }^{-1}\left(\frac{v_{y0}}{v_{x0}}\right) \\ {\theta }_0={\tan }^{-1}\left(\frac{4.0}{9.5}\right) \\ {\theta }_0=22.8° \end{gathered}$$

The range of the athlete without using halters is,

$$\begin{gathered} R_0=\frac{v_0^2\sin 2{\theta }_0}{g} \\ {R}_0=\frac{10.{31}^2\times \sin \left(2\times 22.8°\right)}{9.8} \\ {R}_0=7.75\mathrm{m} \end{gathered}$$

On other hand, if two halters of mass m = 5.50 kg throwing at maximum height, then, by momentum conservation, the subsequent speed of the athlete would be,

$$\begin{gathered} \left(M+2m\right)v_{x0}=Mv{\text{'}}_x \\ v{\text{'}}_x=\frac{\left(M+2m\right)}{M}{v}_{x0} \end{gathered}$$

Thus, change in the xcomponent of the velocity is,

$$\begin{gathered} ∆v_x=v{\text{'}}_x-v_{x0} \\ ∆v_x=\frac{\left(M+2m\right)}{M}{v}_{x0}-v_{x0} \\ ∆v_x=\frac{2m}{M}{v}_{x0} \\ ∆v_x=v{\text{'}}_x-v_{x0} \\ ∆v_x=\frac{\left(M+2m\right)}{M}{v}_{x0}-v_{x0} \\ ∆v_x=\frac{2\times 5.5}{78}\times 9.5 \\ ∆v_x=v{\text{'}}_x-v_{x0} \\ ∆v_x=\frac{\left(M+2m\right)}{M}{v}_{x0}-v_{x0} \\ ∆v_x=1.34\mathrm{m}/\mathrm{s} \end{gathered}$$

The maximum height is attained when,

or,

$v_y=v_{y0}-gt=0or$

$$\begin{gathered} t=\frac{v_{y0}}{g} \\ t=\frac{4.0}{9.8} \\ t=0.41s \end{gathered}$$

Therefore, the increase in the range with use of halters is,

$$\begin{gathered} ∆\mathrm{R}=\left(∆\mathrm{v}{\text{'}}_{\mathrm{x}}\right)\mathrm{t} \\ ∆\mathrm{R}=1.34\times 0.41 \\ ∆\mathrm{R}=55\mathrm{cm} \end{gathered}$$

Hence,the distance that the use of the halters would add to his range is$∆\mathrm{R}=55\mathrm{cm}$ .

Therefore, by using the conservation of momentum, the change in the range can be found.