91Ó°ÊÓ

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

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,$R_0=\frac{v_0^2\sin 2{\mathrm{θ}}_0}{g}............\left(1\right)$

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

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

And take off angle of the athlete is,

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

The range of the athlete without using halters is,

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

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,

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

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,

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

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.