Recent content by FireKyuubi

So is impulse always perpendicular to the surface, like the normal force?

So it's pointed directly up?

Its pointing to the right and up, decreasing vertical velocity and increasing horizontal velocity.

I assumed that the speed would stay constant, but since there's impulse that doesn't have to be the case. There's more horizontal velocity after the bounce than before, is that what you mean?

Oh, sure, I'm assuming that when the ball hits the floor, there will be some impulse that causes the ball's momentum to change. And that impulse divided by time is average force, which should be a constant vector. Taking the magnitude of that constant vector will give me the answer.

##m=.3 kg, v = 11 \frac {m}{s}, t = .25 s, \vec v_1 = mv\langle cos(-55), sin(-55) \rangle, \vec v_2 = mv\langle cos(25), sin(25) \rangle## $$m\vec v_1 - m\vec v_2 = \vec Ft = mv\langle cos(25)-cos(-55), sin(25)-sin(-55) \rangle$$ $$Ft = mv\sqrt{(cos(25)-cos(-55))^2 + (sin(25)-sin(-55))^2}$$ $$F...