How do I Calculat on distance in Catapult project

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To calculate the distance a marshmallow will travel when launched from a catapult, it's essential to consider the elastic potential energy stored in the rubber band, which is not directly calculable with a standard formula. The discussion emphasizes that while the formula Mgh = 1/2MV^2 can be used for velocity, it does not account for the energy in the rubber band, which behaves differently than a spring. Participants suggest that the potential energy of the rubber band must be determined experimentally, as there is no straightforward formula for it. Additionally, energy losses due to the mass of the catapult arm and the gravitational effects when launching must also be considered. Ultimately, practical testing of the catapult design may yield more useful insights than theoretical calculations.
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I'm doing a Assignment is to conststruce a captapult that will launch a marshmallow(Large size) at target on the floor at the other end of the classroom.
I designed catapult. the rules saing the energy;; to launch the marshmallow cannot come directly from the motion of a human being.so i used elastic band .
but, they saing to calculate the distance.
so, i use an energy approach. so I determine the angle at which the mashmallow leaves the catapult.then I assume then energy loss of the arm is gained by the marshmallow.but I ued elastic band, but, i don't know how to measure the load distance to find the energy stored in the elaastic band.how to i take into this measure ments...coz, I don't know the formula of the elasticband energy stored.
just i used this formula to find the velocity.
Mgh = 1/2*M*V^2
but I don't consider the elastice band energy
please anyone could me help to solve this probelm
I need the full foumula how to fine the calculation...
pls sent me to desitamil@gmail.com
Thanks for look through my post.
 
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There is no formula, for that really. You would have to determine the potential energy in the rubber band experimentally.
 
[As I imagine your catapult],
You have a long arm with the marshmallow at the end. The arm is pulled back and the rubber band is stretched out. When the arm is released, the arm and marshmallow fling forward [and ideally hit your target].

When you pull back the arm of the catapult and stretch the rubber band, you are storing elastic potential energy in the rubber band. You assume that it is all converted into kinetic energy of marshmallow as it leave the arm in the end.
The formula for the elastic potential energy of a spring (not a rubber band) is,
PE = 1/2*k*x^2, where k is the spring constant and x is the distance the spring is compressed/expanded from its normal state. Although a rubber band isn’t exactly a spring, and you don’t know the spring constant, the formula might still be of some use to you to give you an idea.

All the energy stored in the rubber band is not necessarily converted into kinetic energy of the marshmallow though. By comparison to the arm of the catapult, the marshmallow has a low mass (after all, it is a marshmallow). Energy had to go into making the arm of the catapult move as well, and if your arm is significantly more massive (and long), it might take a whole lot of energy away from the marshmallow.
Also, assuming the arm is pulled back horizontal, when released, it must raise itself and the marshmallow up (against gravity), which takes energy.


I am probably thinking way too much into such an assignment, it would probably be much more practical to test over and over again your design rather than worry about calculating energy and trajectories.
 
hi mrjeffy
tx for u r advice...
i got some idea from u r post...i think I don't care abt potential energy of the rubber band ,if we assuming the arm is pulled back horizontal, when released isn't it?...so, Mgh = 1/2MV^2 is the only formula to fine velocity...rite..
 
You need to care about PE of the rubber band if it is what is going to propel the item to the target. Don't forget, the PE of an object is not necessarily PE=mgh. Think of the rubber band as a spring. What is the PE of a spring when extended from it's equilibrium point? Hint: it is not mgh.
 
colombo said:
hi mrjeffy
tx for u r advice...
i got some idea from u r post...i think I don't care abt potential energy of the rubber band ,if we assuming the arm is pulled back horizontal, when released isn't it?...so, Mgh = 1/2MV^2 is the only formula to fine velocity...rite..

Wrong.

As I said above, there is no formula for your setup. There is no formula for the potential energy of a rubber band, it is not mgh.

You cannot use conservation of energy without knowing the potential energy in the rubber band. And there is no formula for that. a rubber band is not a simple spring.
 
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