The following occurs in 6 seconds (6 points on the graph):

Starting at y = 0 m, a stone is thrown straight up on the edge of a cliff until it reaches its maximum height at 2 seconds, and begins to fall. At 3 seconds, it is level with the 1 second point. At 4 seconds, the stone is level with y = 0 m, where it continues downward for another 2 seconds, ending at the sixth second. My displacement graph was correct, but my velocity graph is really confusing. In particular I am having trouble finding how exactly to plot this.

Things I figured out, but don't know if it's correct:

a. Going up 2 seconds is the same as free falling for 2 seconds? So the initial velocity is +19.6 m/s.

b. At the 1 second mark, the velocity is the (free-fall value of t=2) - (free-fall value of t=1) = +14.7 m/s

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c. The velocity at the 2 second mark is 0.

d. At the 3 second mark, the velocity is the free-fall value of t=1. So 3 = -4.9 m/s.

e. At the 4 second mark, the velocity is the (free-fall value of t=2)/2. -9.8 m/s

f. At 5 seconds, (free-fall value of t=3)/3. v = -14.7 m/s

g. At 6 seconds, (free-fall value of t=4)/4. v = -19.6 m/s

I know that something or everything is wrong, but I can't seem to think straight at the moment.