Question

In the Olympiad of 708 B.C. 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 the flight. Suppose a modern 79 kg long jumper similarly uses two 4.50 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 vvec = (9.5ihat + 4.0jhat) m/s with or without the halteres, and assume that he lands at the liftoff level. What distance would the use of the halteres add to his range?

Answer 1

0.441296016722 m

Step-by-step explanation:

`v_0` = Initial speed =
`9.5\hat{i}+4\hat{j}`

The magnitude of velocity is

`v_0=√(9.5^2+4^2)\\\Rightarrow v_0=10.307764064\ m/s`

Take of angle

`\theta=tan^(-1)(4)/(9.5)=22.833\ ^(\circ)`

Range without halters

`R=(v_0sin2\theta)/(g)\\\Rightarrow R=(10.307764064^2sin(2* 22.833))/(9.81)\\\Rightarrow R=7.74\ m`

In this system the momentum is conserved

`(M+2m)v_x=Mv_x'\\\Rightarrow v_x'=((M+2m)v_x)/(M)`

Change in velocity is given by

`\Delta v_x=v_x'-v_x\\\Rightarrow \Delta v_x=((M+2m)v_x)/(M)-v_x\\\Rightarrow \Delta v_x=(2m)/(M)v_x\\\Rightarrow \Delta v_x=(2* 4.5)/(79)* 9.5\\\Rightarrow \Delta v_x=1.08227848101\ m/s`

Time taken is

`t=(v_y)/(g)\\\Rightarrow t=(4)/(9.81)\\\Rightarrow t=0.407747196738\ s`

Difference in range is given by

`\Delta R=\Delta v_xt\\\Rightarrow \Delta R=1.08227848101* 0.407747196738\\\Rightarrow \Delta R=0.441296016722\ m`

The difference in range is 0.441296016722 m