Yes to the first question. They must be talking about motion along one axis. You can go backwards for a while (negative velocity) and still get you your destination.
As for the second question
Y1 = H - (g/2)t^2
Vf = sqrt (2gH)
Y2 = Vf*t - (g/2)t^2
When Y1 = Y2,
Vf*t = H
Vf^2 t^2 = H^2
2 g H t^2 = H^2
gt^2 = H/2
Y1 = Y2 = 3H/4
Yes, you are right, and your reasoning led you to the correct answer right away
The average velocity for a trip has a positive value. Is it possible for the instantaneous velocity at any point during the trip to have a negative value? (I'm thinking yes because derivatives can be negative but I'm not sure).
also
A ball is dropped from rest from the top of a building and strikes the ground with a speed Vf. From ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is Vo=Vf, the same speed with which the first ball will eventually strike the ground. Ignoring air resistance, decide whether the balls cross paths at half the height of the building, above the halfway point, or below the halfway point. Give your reasoning. (I'm thinking the second ball must be traveling faster if it starts at the fastest speed of the first ball, so they'd cross above the halfway point. Is that right?)
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