1. The problem statement, all variables and given/known data A kangaroo jumps a height of 2.8m. How long was it in the air? 2. Relevant equations v=v0+at x=x0+v0t+(1/2)at^2 v^2=v0^2+2aΔx 3. The attempt at a solution -2.8 = -4.9t^2 .57 = t^2 t = .76 s (down) 0 = v0^2+2(9.8)(2.8) 54.88 = v0^2 v0 = 7.408 0 = 7.408-9.8t t = .76 (up) t = 1.5 sec SO... I know I have the answer right I just don't understand how that works conceptually - that if something has a initial velocity upwards it takes the same time as if the initial velocity was 0 downwards. How can something traveling the same distance (just one negative and one positive) with the same acceleration but different initial velocities take the same amount of time out to the 9th decimal??