Homework help: Dropping a sand bag from a Hot Air Balloon

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When dropping a sandbag from a hot air balloon, the initial velocity (v0) of the balloon must be considered in the free fall equation. If the balloon is rising rapidly, the sandbag will have the same upward velocity as the balloon at the moment of release, affecting its trajectory. This contrasts with dropping the sandbag from a stationary position, where it would fall directly downward with zero initial velocity. The acceleration due to gravity remains constant, but the initial conditions lead to different fall paths. Understanding these dynamics is crucial for accurate calculations in physics.
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Homework Statement
Suppose an air baloon keep raising with constant velocity v0, at time t=0 the sand bag attached to it drop at rest.
Relevant Equations
y=vo-1/2gt^2
In this situation should my free fall equation contain the v0 of the baloon or I should deny it. Because it seems to me that there is no outer force acts on the sandbag, so the scenario is just the same as I climb to the same height at time t=0 and drop the sandbag at rest.
 
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Clockclocle said:
so the scenario is just the same as I climb to the same height at time t=0 and drop the sandbag at rest.
What if the balloon is rising at the speed of a rifle bullet? You are riding on the balloon holding the sandbag in your hand and let it go.
 
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I see the mistake, I thought that it gonna fall below me so it would be the same when I stand still. But in this case I keep moving with velocity v0.
 
Clockclocle said:
##\dots## so the scenario is just the same as I climb to the same height at time t=0 and drop the sandbag at rest.
The acceleration is the same not the scenario. In the first case the velocity of the bag relative to the ground is the same as that of the balloon whilst in the second case it is zero. Different initial velocities mean different free fall trajectories.
 
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