- #1

Milchstrabe

- 74

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At first glance the physics of a seesaw seem quite simple, however I've been stumped on a problem. For part of my Physics project, I am incorporating a seesaw catapult where one weight will be dropped on one side launching the hacky sack on the other side. Simple right? Of course... but the hacky sack needs to land .76 m to the left, on a stool .76 m high.

I figured my launch angle to be approx... 80 degrees ( I can always change this) which would yeild me a 5.45 m/s vertical velocity and a .8 (approx) m/s horizontal velocity. The Tanget velocity would therefore need to be 5.55 m/s (approx). Now how can I work backwards to solve for how heavy and at what height the object would have to be dropped from in order to launch it at that velocity? Let's say the arm with the hacky sack is 4 times larger then the other arm, so we have 4 and 1. I'm guessing I'll have to use Torque, and force of course.

I gave it some thought tho, and in order for the hacky to even move the mass falling must have enough force to bring the seesaw to equilibrium. So once I calculate that force required, then anything more than that will begin to bring the acceleration and velocity of the hacky up from zero. Also I must find a way to factor in the weight of the arms, it will make a difference. Any help guys?! Sorry if it was confusing