posted by john .
During liftoff, a hot-air balloon accelerates upward at a rate of 3.4 m/s^2 . The balloonist drops an object over the side of the gondola when the speed is 18 m/s.
What is the magnitude of the object’s acceleration after it is released (relative to the ground)?
You need to know how much mass was dropped, compared to the mass of the balloon. The buoyancy force will be the same, but the acceleration will increase in proportion to the reduction of mass.
Knowing the speed of the balloon when the mass was dropped does not provide the needed information.
i found the answer to the question it is 9.8 m/s^2. but how long would it take to hit the ground?
You are right. I misread the problem. I thought they were asking for the acceleration rate of the balloon after the object was released. A 'duh' moment.
Once released, the dropped object accelerates downward at the acceleration of gravity, g = 9.8 m/s^2
The time after liftoff when the object was released was:
t = V/a = 18/3.4 = 5.3 seconds.
The height it had risen at that time was
(1/2)at^2 = 47.6 m
Now you can calculate the additional time t' it takes for the dropped object to hit the ground.
y = 47.6 + 18 t' - 4.9 t'^2 = 0
Solve that quadratic for t' and take the positive root.
thank you very much