Recent content by Dark_Rak3r

Thanks for the help. I'll definitely come back to these boards in the future if I need anything else. EDIT: AAAARRRRRRGGGGGGG! I just realized that I had figured out the time equation earlier without realizing it was the right equation. I took the equation h=vi∆t+(1/2)g(t)2 and figured out that...

So t=√(2h/g) and the final equation is ∆R=√(kx2/m)*√(2h/g)

So, just to keep the variables similar, the equation could be written as: h=hi+vt-(1/2)gt2 where all variables correspond to those in my first post? Also, if h is the vertical displacement, can I use: ∆d=vi∆t+1/2a(∆t)2 where vi∆t=0 because the initial vertical velocity is 0, and the rest can be...

I see what you mean by the displacement being 0. What does s represent in the equation you gave? The sine of the launch angle?

Why would the vertical displacement be zero? The ball is assumed to be starting above the ground, in which case the vertical displacement would be a non-zero number (again, the ball is being shot at a perfectly flat angle, only completing one half of a parabola). Your math is also kind of going...

I would, but I need to somehow come up with a final equation using only those variables.

Homework Statement A spring is pulled back and launches a ball. The ball flies into the air at a perfectly horizontal angle. How far will the ball go before it hits the ground? Variables for the FINAL equation are: ∆x - distance the spring is pulled back g - acceleration of gravity m - mass of...