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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 ?

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can u elaborate "angle of the drag" and "distance the sling was dragged"?? –  Vineet Menon Oct 11 '11 at 6:34
    
Basically from dragging I mean change in the position of the slingshot stone i.e. moving it backwards in order to give it speed and a angle for launch. –  y ramesh rao Oct 11 '11 at 7:45
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2 Answers

up vote 1 down vote accepted

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,

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so is the hookes law coefficient dependent on the type of the material we are using for creating the sling ? –  y ramesh rao Oct 11 '11 at 13:58
    
true...it depends on so many things...one of them is the material.. –  Vineet Menon Oct 11 '11 at 17:07
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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.

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