Mass Effect on Slingshot Motion? For my physics class (I'm a high school student), we created slingshots. Our task is to predict the distance a projectile, launched from a slingshot using surgical tubing, would land. We aren't given the mass of the projectile until the day of testing, so we would have to use a graph to find out the distance. My partner and I conducted a series of tests comparing mass to distance from a given pullback length ($55$ cm). The angle the projectile is launching from is $54^\circ$. K is not linear, but for our set pullback length our K value is $100$ (N/m). Our results show a cubic, or possibly quadratic graph. Why is this? Why would the relationship between mass launched and distance follow this shape?  Here is a graph of our data:

 A: I congratulate your teacher on posing a fascinating intellectual challenge. So here's my guess (as a former physics student turned (medical) physician) constructed by analogy with the typical analysis of simple harmonic motion. In that situation with a spring constant of K and the force dependent on instantaneous displacement from the center of the oscillation, there is a resonant frequency that is a function of the mass and the spring constant. You are essentially looking at one-half of such an oscillator, so I think there would be an optimal ratio of mass to spring tension that would achieve the maximal transfer of energy from tubing to projectile. The resonant frequency of such a system would be one half the inverse of the time from release to the midpoint of the slingshot. (This of course assumes you can measure the time from release to crossing the slingshot "Y", but I'm sure you have some high-speed cameras around for that task, right?)
Do some searching on this website or on Wikipedia for the calculation of resonant frequency in "simple harmonic motion" and measure the spring constant of your slingshot. See if you can pull all these together for an A+ scientific report describing your experimental results!
A: The calculations are going a bit long, but I have an approximate explanation, Just include viscous force given by stokes law in the equations of parabolic/projectile motion and then calculate the distance/range of projectile. Here goes : 

At any time $t$ the viscous force on the projectile will be $6\pi\eta a v$ if the radius of "spherical" ball is $a$ and it is moving with $v$ velocity.
  Note : $\eta$ is the viscosity of fluid/air the ball moves in.
  Breaking force into components,  
$F_x = {-6}\pi\eta a v \cos\theta$
  $F_y = {-6}\pi\eta a v \sin\theta$
  ${d{v_x}}/{dt} = (({-6}\pi\eta a {v_x})/m) dt$
  ${d{v_y}}/{dt} = (({-6}\pi\eta a {v_y} + mg)/m) dt$
  $v_x = u \cos\theta e^{({-6}\pi\eta a t)/m}$
  $v_y = ((mg + 6\pi\eta a u \sin\theta) e^{({-6}\pi\eta a t)/m} - mg)/(6 \pi\eta a)$  
$dx = v_x dt$ on integrating:
  $R = (m u \cos\theta)/({6 \pi \eta a}) × [ 1 - e^{({-6}\pi\eta a T)/m}]$ T is time taken to cover range.  
Assuming you get at $t = T$, $v_y = -u \sin\theta$, then putting value $v_y = - u sin\theta$ in equation for $v_y$ solving for $t$ and putting in equation of $R$, we get,
  $ R = (m u^2 \sin{2\theta})/(mg + 6 \pi\eta a u \sin\theta)$  
Now since you are using a spring to launch the projectile, we have
  $K x^2 = m u^2$
  $R = K x^2 \sin{2\theta} [\sqrt[2]{m} / (m^{3/2} g + 6 \pi\eta a x \sqrt[2]{K} \sin\theta) ] $  
This implies
  $ R \propto 1/(mg + A/{\sqrt[2]{m}}) $  
If you differentiate $R$ with respect to $m$ you will find that a maxima is achieved when $ m = A/(2 m^{3/2}) $
  Here $ A = 6 \pi\eta a x \sqrt[2]{K} \sin\theta)$

Note :
1.  I have assumed that while landing $v_y = {- u \sin\theta}$ This maybe a close approximation but it is not exact, to find exact equation you have to integrate for height the same way I have done for range and then find at what times it is zero, since the equations are far too complex I have omitted them and taken approximations.
2. I have assumed projectiles to be moving in a fluid i.e. air.
3. I have also assumed air to be simply present, not moving.
A: Assuming throwing force and launch angle were the same for all projectiles, the answer is probably much simpler than DWin's.   
The effect of the air resistance on the lighter projectiles was greater than fore the heavier ones, whereas the heavier objects fell short due to lower speed in the first place, because the throwing force was not able to accelerate them as much as the lighter ones.
