Slingshot Projectile motion calculation If I have distance the sling was dragged and the angle of the drag; then I require to calculate the projectile motion this sling shot would make.
What are the equations that would enable this calculation?
 A: see this page
$$ y=y_0 + x tan\theta -\frac{gx^2}{2(vcos\theta)^2}$$
assuming $y_0$ to be 0 the equation becomes,
$$ y=x tan\theta -\frac{gx^2}{2(vcos\theta)^2}$$
Now, the only variable what you need are $v$ and $\theta$, among which you have $\theta$.
So, you need to convert your "distance dragged" to $v$.
Energy stored in the sling just before shooting will be $$U=kx^2$$ where $k$ is Hooke's constant and x is your "distance dragged".
so, 
$$K.E. = U = kx^2$$
$$\frac{1}{2} mv^2 = kx^2$$
$$v=\sqrt{\frac{2kx^2}{m}}$$
Hope this might help.
Regards,
A: I had a similar school science project a long time ago.  Fortunately, this project has left me notes.   Let me share an interesting  case where there is a closed form solution:  
This is the projectile motion on a flat trajectory when the resistance of the medium(air) is proportional to the square of the velocity. The flat trajectory means that a projectile is launched at angles $\theta_0<15^{\circ}$. The derivation process is unfortunately too long. So I'll get away with just the final result, the equation of the trajectory.  
Air resistance: $F=kv^2$
Initial velocity:$v_0$
Initial angle:$\theta_0 $
Acceleration of gravity:$g$
The approximate equation of the trajectory of a projectile:  
$$y=x\tan\theta_0-\frac{g}{(2kv_0)^2}(e^{2\mu}-2\mu-1)$$  
where $\mu=\frac{kx}{\cos\theta_0}$  
To compare the trajectory in air with a trajectory in a vacuum, expand $e^{2\mu}$ into Taylor series and after obvious cancellations we get:  
$$y=x\tan\theta_0-\frac{g}{2}\left(\frac{x}{v_0\cos\theta_0}\right)^2-\frac{gkx}{3}\left(\frac{x}{v_0\cos\theta_0}\right)^2-...$$  
Here the collection of the first two terms, independent of the drag coefficient $k$, coincides with the equation of the trajectory of the projectile in a vacuum, the third term gives the correction due to the impact of the resistance: That means, the actual trajectory is below the parabola.
