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**1. Homework Statement**

"In the Olympiad, some athletes competing in the standing long jump used handheld weights called

*halteres*to lengthen their jumps. The weights were swung up in front just before liftoff and then swung down and thrown backward during flight. Suppose a modern ##78kg## long jumper similarly uses two ##\frac{11}{2}kg## halteres, 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 ##v=\langle\frac{19}{2},4\rangle\frac{m}{s}## with or without the halteres and assume that he lands at the liftoff level. What distance would the use of halteres add to his range?

**2. Homework Equations**

##m_0=78kg##

##m_1=11kg##

##v_0=\langle\frac{19}{2},4\rangle\frac{m}{s}##

##x=(\frac{||v||^2}{g})(sin2\theta_0)##

##\theta_0=tan^{-1}(\frac{8}{19})##

**3. The Attempt at a Solution**

##p_0=(78kg)\langle\frac{19}{2},4\rangle\frac{m}{s}##

##Δp=(11kg)\langle0,4\rangle\frac{m}{s}##

##p_f=p_0+Δp##

##||v_0||=||\frac{p_0}{m_0}||##

##p_f=m_0p_0+m_1Δp##

##(m_0+m_1)v_f=m_0p_0+m_1Δp##

##v_f=\frac{m_0p_0+m_1Δp}{m_0+m_1}##

##||v_f||=\frac{1}{m_0+m_1}||m_0p_0+m_1Δp||##

I'm thinking that since the x-distance from the starting point and the point of maximum height is the same as the distance from the vertex to the landing spot, the x-distance between the maximum heights of the trajectories is the same as the x-distance between their landing spots. Forgive me; the FBD I drew on paint isn't to scale. Anyway, the final answer would come out to:

##Δx=(\frac{sin2\theta_0}{g})(||v_f||^2-||v_0||^2)##

Only problem, is that: ##||v_f||<||v_0||##. So I'm thinking that I messed up around this point.

##p_f=m_0p_0+m_1Δp##

##(m_0+m_1)v_f=m_0p_0+m_1Δp##

##v_f=\frac{m_0p_0+m_1Δp}{m_0+m_1}##

##||v_f||=(\frac{1}{m_0+m_1})||m_0p_0+m_1Δp||##

Can anyone tell me what is wrong with the relations that I used?

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