Going back to the energy equations, I think you missed an 'm'. It's supposed to say -2mgy right? And for the y, would that be the distance above ground at the end? So wouldn't that be 0, making Eg negligible?
So setting y0 = -0.48 I derived the following equation:
y = -0.48 + xsinθ/vcosθ - 1/2g(x/vcosθ)^2
Is this correct? Now all I have to do is set the equation to zero, and plug in the x which is the given distance, and solve for v. Then from there I work backwards and use the energy eq'ns to...
I understand what you're saying but shouldn't the ground be -0.48m? The distance from the ground to where the marble is launched on my catapult is 48cm above ground level. Or is this distance insignificant enough to rule out?
Ok so once I have the coefficient, I will use your derived equation to solve for velocity. Then how do I find the specific range? would I use this equation:
ΔDx = (V1sin2θ)/g
Actually nevermind my previous post. Would I use the following equation:
Elastic energy = gravitational potential + kinetic energy
Ee = Eg + Ek
If so how would I calculate the elastic constant for my specific bungee cord?
My current physics project is to design a catapult that will launch a marble to a given distance (between 2-3m). I have successfully made one out of wood, using a bungee cord to snap the arm back after being pulled on.
My question is how do I calibrate this properly? Our teacher said we need...