# How would I use momentum to calculate projectile range?

1. Nov 12, 2016

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"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. Relevant 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?

Last edited: Nov 12, 2016
2. Nov 12, 2016

### Eclair_de_XII

Hold on, I forgot something: If the two halteres are moving $0\frac{m}{s}$ relative to the ground, then I can assume that the long jumper throws them back with a velocity of $-v_{0x}$.

3. Nov 12, 2016

### TomHart

Edit: Yeah, the quote kind of butchered it.

I don't know if the quote (above) will come out the same or not. In your initial momentum equation, it looks like you didn't include the mass of the two halterers. Also, I'm not familiar with your notation. Does the lift off velocity indicate that the x component is 9.5 m/s and the y component is 4 m/s?
Edit changed 8.5 m/s to 9.5 m/s

4. Nov 12, 2016

### Eclair_de_XII

That's the momentum if the the halteres were not included. And yes.

5. Nov 12, 2016

### TomHart

Doesn't he have them in his hands when he lifts off?

6. Nov 12, 2016

### Eclair_de_XII

You're right. In any case, I found the answer already. This is the equation describing the act of throwing the halteres.

$m_1v_{0x}=m_0(v_{0x}+(\frac{m_1}{m_0})v_{0x})=m_1v_{0x}=m_0(v_{0x})(1+\frac{m_1}{m_0})$

These are the equations to find the difference in horizontal range had the long jumper not used the halteres.

$x_0=(\frac{||v_0||^2}{g})(sin2\theta_0)$
$x_f=(\frac{||v_0||^2}{g})(sin2\theta_0)+(v_{0x})(\frac{m_1}{m_0})t$
$Δx=x_f-x_0=(v_{0x})(\frac{m_1}{m_0})t$

This is to find the unknown $t$ it takes for the jumper to land.

$0=v_{0y}-gt$
$t=\frac{v_{0y}}{g}$

Substitute in, and--

$Δx=(v_{0x})(\frac{m_1}{m_0})t=(v_{0x})(\frac{m_1}{m_0})(\frac{v_{0y}}{g})=0.55m=55cm$

Someone correct me if I am mistaken in my thought process.