physics
posted by Anonymous .
During liftoff, a hotair balloon accelerates upward at a rate of 3.3m/s^2. The balloonist drops an object over the side of the gondola when the speed is 15m/s.
Part A: What is the magnitude of the objectâ€™s acceleration after it is released (relative to the ground)?
Part B: How long does it take to hit the ground?

F = m a
m g is the only force on the object
m g = m a
a = g = 9.81 m/s^2
now where is the balloon?
v = 0 + 3.3 t
15 = 3.3 t
t = 4.55 seconds to reach v = 15
h = (1/2)a t^2 = (1/2)(3.3)(4.55^2) = 34.2 m high
so we release the rock at 34.2 meters while moving up at 15 m/s
a = 9.81
h = 34.2 + 15 t  4.9 t^2
when is h = 0 ? (hits ground)
4.9 t^2 15 t  34.2 = 0
t = [ 15 +/  sqrt (225 + 670) ] /9.8
t = [ 15 + 29.9 ]/9.8
t = 4.58 seconds