# What are all the variables that need to be taken into account when recording the time of flight and distance of a projectile fired from a catapult (tension, angle, mass of projectile, etc)?

Jul 18, 2017

Assuming no air resistance (reasonable at low velocity for a small, dense projectile) it isn't too complex.

#### Explanation:

I'm assuming that you are happy with Donatello's modification to / clarification of your question.

The maximum range is given by firing at 45 degrees to the horizontal.

All the energy provided by the catapult is expended against gravity, so we can say that the energy stored in the elastic is equal to the potential energy gained. So E(e) = $\frac{1}{2} k . {x}^{2}$ = m.g.h

You find k (Hooke's constant) by measuring the extension given a load on the elastic (F=k.x), measure the extension used to launch and the mass of the projectile and can then get the height it will rise to, if fired vertically.

The time of flight is independent of the angle, as the projectile is in free fall from the moment it leaves the catapult, irrespective of how it is launched. Knowing the initial elastic energy (called E(e) above) you can find it's initial velocity, u from E(e) = $\frac{1}{2.} m . {u}^{2}$ and then the time of flight by substitution into v = u + a.t where v is the final velocity (zero) at the maximum height. The total time of flight will be double this, once when rising, once when falling.

Finally, you can calculate range, R from R = $\frac{{u}^{2.} \sin \left(\theta\right)}{g}$