Recent content by jackneedshelp

Ohhh kaaaaay! I get it! Thanks!

d = vi t + (0.5)(a)(t)sqrd 2 = sqrt(19.62h) t + 0 t = 2 / sqrt(19.62h)

vf srd = vi srd + 2ad vf srd = (sqrt(19.62h))srd + 2(0)(2) vf = sqrt(19.62h) so vf = vi

d = 2.0m vi = sqrt(19.62h) vf = 0 t = ? 2 = (0+sqrt(19.62h))/2 *t t = 4 / (sqrt(19.62h)) is that what I'm supposed to do?

or final velocity?

so I'm solving for time?

so I did the horizontal part, and got the horizontal acceleration to be 4.905h. how do I do the vertical part?

I set potential energy equal to kinetic energy, and solved for v. v = sqrt(19.62h)

I don't even know where to start! It should have something to do with momentum and impulse, but I can't even get started on it! Please help!

Homework Statement A 0.010 kg pendulum bob is dropped from a height (h) above its equilibrium position. When the bob reaches its equilibrium position, the string breaks and the bob now acts as a projectile. After the string breaks the bob falls 1.5 m while moving 2.0 m horizontally...