posted by jeff .
You are in a hot-air balloon that ascends at a constant velocity of 2.00 m/s. When the ballon is 1000 meters in the air you jump out of it what speed will you hit the ground with? neglect air resistance.
I know we use the conservation of mechanical energy to solve this my question is will my initial velocity be 2.00 m/s or will it be zero because i am at rest in the ballon.
Technically, when you detach yourself from the balloon, you have an initial upward velocity of 2m/s which, after 0.20sec, you will rise .204m.
Ignoring this, the time to reach the ground derives from h = Vot + gt^2/2 or 1000 = o(t) + 9.8t^2/2 yielding a falling time of 14.28 sec.
The velocity with which you hit the ground derives from Vf = Vo + 9.8t = 0 + 9.8(14.28) = 140m/s.
The "exact" answer combines the two results.