64th

Problem ONE: The Evil One, The Throwing Up One. So, I realize there are actually two steps to this one. Finding all the unknowns for the peak and then going back down again. With THAT said... Step I - vi: 25m/s vf: 0m/s a: -9.81m/s^2 d: ? t: ? So, I tried two things with T. The first equation I used initially in Problem TWO, and came up with: t: 0.3924s And then I did it again, this time with... t= vf - vi / a t= -25m/s / -9.81m/s^2 t= 2.5484s Okay. What? Wait... No no... No. D: And yet, I proceed with the first equation. d= di + vi(t) + 1/2 at^2 d= 0 + (25m/s)(.3924s) + 1/2 (-9.81m/s^2)(.3924s^2) d= 9.81 - .755 d= 9.055m And then with the second. d= di + vi(t) + 1/2 at^2 d= 0 + 25m/s(2.55s) + 1/2(-9.81m/s^2)(2.55s^2) d= 63.75 - 31.89475 d= 31.855m D: Is it possible for a ball to travel upwards 31 meters in under 3 seconds? Somehow.... I have my reservations. Right. So afterwards I proceeded to turn myself around in circles. If you really want the confused math, I guess I could post it all. But mostly, if you guys can provide me with a formula that will give me a decent answer... I can finish it off. I believe I used the freefall equation thing- d: .5(9.81m/s^2)(t^2) To do the chart. Or would that be cheating? Probably... Google isn't helping much either. :/

Oh god. u.u I'll try as best I can to organize my thoughts then. Problem TWO: The guy throwing the ball down from 15 meters. Okay, easy. I have a handy formula for this. Find the 'knowns': vi: 0m/s vf: ? a: +9.81m/s^2 d: 15m t: ? t= + sqrt[0-2(9.81m/s^2)(15m)]/9.81m/s^2 t= 1.75s vf= vi + at vf= 0 + (9.81m/s^2)(1.75s) vf= 17.17m/s Then I did a little chart of it, manually, detailing the displacement after every .25s: .25s = .318825 m .50s = 1.226525 m .75s = 3.67875 m (I don't think this is correct) 1.00s = 4.905 m 1.25s = 7.6640625 m 1.50s = 11.03625 m 1.75s = 15.0215625 m So that problem checked out okay. I'll be really heartbroken if it's wrong.

So the question is horribly vague and we were told to, by any means, figure it out. I'm begging all of you, to put me out of my misery here. --- A ball is thrown up into the air at an upward velocity of 25m/s, consequently, a ball is thrown down towards the ground from a building 15 meters high. At what point do the two balls meet, and (I'm assuming here) at what hieght? --- Again, no details were given in reference to the force thrown, or displacement. Say the ball dropped from the building has an initial velocity of 0m/s to make things a bit easier. And say we don't take into consideration how tall the person throwing the ball is. That it is, mysteriously, just moving off of the ground on it's own accord. Any help given would be GREATLY appreciated, as this stupid question has killed my weekend. (Though, the second of the two problems is pretty much solved).