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

  • Thread starter Thread starter Clockclocle
  • Start date Start date
  • Tags Tags
    equation Free fall
AI Thread Summary
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.
Clockclocle
Messages
31
Reaction score
1
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.
 
Physics news on Phys.org
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.
 
  • Like
Likes Clockclocle
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.
 
Last edited:
  • Love
  • Like
Likes MatinSAR and Delta2
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top