Recent content by groetschel

You mean something like this???

Then I have no idea how to solve this... Could you give me a hint?

Ok, let me try this... ##h - vt - \frac 1 2 g t^2 = 0## $$ ax^2 + bx + c = 0 $$

My result is 1.19s 1.19s does not make sense, because B will travel D = v*t = 5.833 m/s * 1.19s = 6.9 m This does not make sense, because the distance between A and B is only 2 m.

??? Please try it on your own: H = 2m v = 5.833 m/s g = 9.8 m/s2 Solve the equation for t...

2m - 1/2*9.8m/s2*t2=5.833m/s2*t 1/2*9.8m/s2 * t2 + 5.833m/s * t = 2m t2 + 2*5.833m/s /9.8m/s2 * t = 4m t (t + 1.19s)2 = 4m Something is wrong, isn't it?

Yeah, you are right. the units of the velocity are incorrect - I corrected them. I drew a diagram, but what is wrong? The acceleration works downwards to earth, the speed upwards... Where can I go wrong?

A: Distance = 2m - 1/2*9.8m/s2*t2 B: Distance = 5.833m/s*t A: Distance = B: Distance 2m - 1/2*9.8m/s2*t2=5.833m/s*t The question is now: how do I solve this quadratic function? All my attempts do not make sense.

Yes, I would say so. B climbs: positive, A falls: negative.

B starts at y=0 straight up A starts at a height of 2m directly above B and falls with 9.8m/s2

No, I am sorry. What I have given is all I have. In the end, it is just a quadratic function I have to solve for t, but my result is only crap.

DofA = D0 + 1/2at^2 DofB = vt DofA = DofB, but how do I solve this and get the time when A and B hit?