A stone is thrown upward from the edge of a cliff, reaches its maximum height, and then falls down into the valley below. A motion diagram for this situation is given View Figure https://www.physicsforums.com/attachment.php?attachmentid=6246&stc=1&d=1139278282 , beginning the instant the stone leaves the throwerâ€™s hand. Construct the corresponding motion graphs. Ignore air resistance. In all three motion graphs, one unit of time is equivalent to one unit of time (one second) in the given motion diagram. Construct a graph corresponding to the stone's vertical displacement, y(t). Ok, I know that the intial position is 0. However, when I want to find the displacement value at t = 2... I do this: I use Vf = Vo + at to find the intial velocity thrown above.. I use -9.8 for a, t for 2, vf for 0 since it's at the top. Then I plug the Vo into Vf^2 = Vo^2 + 2as, and solve for s. However, I get 19.6 and that isn't within bounds for me to draw the graph.. What am I doing wrong? or Could I use another technique? Thanks
Uniform acceleration, -9.8 m/s^{2}, i.e. acceleration is downward. Two equations: y(t) = y_{o} + 1/2 a t^{2} gives y(t), neglecting wind resistance. v(t) = v_{o} + at