Calculating Weight to Make Balloon Rise 116m in 17s Using Archimedes Principle

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To calculate the weight needed to make a lighter-than-air balloon rise 116 meters in 17 seconds using Archimedes' Principle, the buoyant force must be determined first. The buoyant force equals the weight of the air displaced by the balloon, which can be calculated using the balloon's volume and the density of air. The resultant force, which causes the balloon to rise, is the difference between the buoyant force and the weight of the balloon. To find the initial mass of the balloon, the forces must be in equilibrium, meaning the weight of the balloon equals the buoyant force when stationary. The discussion emphasizes the need to calculate the initial mass to determine how much ballast must be dropped to achieve the desired ascent.
moeraeleizhaj
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A lighter-than-air spherical balloon and its load of passengers and ballast are floating stationary above the earth. Ballast is weight (of negligible volume) that can be dropped overboard to make the balloon rise. The radius of this balloon is 6.59 m. Assuming a constant value of 1.29 kg/m3 for the density of air, determine how much weight must be dropped overboard to make the balloon rise 116 m in 17.0 s.

Hi, that's a physics problem aht confounds me.
It's about archimedes principle (weight of fluid displaced=bouyant force) , right? I've found out the bouyant force using that principle.

And the part when the baloon moves can be solved by doing (bouyant force-weight of the balloon=Resultant force) wherein the resultant force causes the balloon to rise up(resultant force=mass of balloon*vel (116-17)). Is that assumption right? if i do that i get the new mass of the balloon but i don't know how to get it's intial mass and i need it to find out how much mass needs to be dropped.

Can anyone help me with my problem please?
 
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moeraeleizhaj said:
A lighter-than-air spherical balloon and its load of passengers and ballast are floating stationary above the earth.
What can you deduce from this statement?
 
that the forces are in equlibruim .'. the weight of the balloon=the bouyant force?
 
moeraeleizhaj said:
that the forces are in equlibruim .'. the weight of the balloon=the bouyant force?
Exactly.
 
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 .
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