During liftoff, a hot-air 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.

Question

Part A: What is the magnitude of the object’s acceleration after it is released (relative to the ground)?

Part B: How long does it take to hit the ground?

Answer 1

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