I made a guess based on the sig-figs of the other numbers and used 1.80 (instead of 1.84) and I got it right. This is going to be a long semester if more stuff like this happens.
My professor is doing me a favor because I turned it in early, he said I can still try one more time to get it right. Ill try the more accurate gravity constant. But ill make sure to figure out what went wrong and post it.
Alternative formulae
Yes.
Update:
Using
vf = vi +at2
and
d = ((vi + vf)/2) *t
I have calculated displacement again getting the same answer.
Edit: I think its time to move on. When I get the solution from the professor or a student who got it right, ill post it here for any future...
Math fun
I thought maybe my mistake was that I should calculate as t=0 for first stone, and t=4 for the second stone (t and t+4). I thought wrong.
vt+(-\frac{g}{2})t2=v(t-4)+(-g)(t-4)2
vt - \frac{g}{2}t2=vt - 4v - \frac{g}{2}t2 + 4gt - 8g //expand and distribute.
0 = -4v + 4gt -8g...
Homework Statement
Suppose you throw a stone straight up with an initial velocity of 20.5 m/s and, 4.0 s later you throw a second stone straight up with the same initial velocity. The first stone going down will meet the second stone going up. At what height above the point of release do the...