Question

A 5.45-m radius air balloon loaded with passengers and ballast is floating at a fixed altitude. Determine how much weight (ballast) must be dropped overboard to make the balloon rise 106 m in 21.5 s. Assume a constant value of 1.2 kg/m3 for the density of air. Ballast is weight of negligible volume that can be dropped overboard to make the balloon rise.

Answer 1

Mass of the balloon is 45.5 kg.

Mass is the intrinsic property of any object.

Radius of balloon, r = 5.45 m

Distance moved by the balloon, d = 106 m

Time spent in moving, t = 21.5 s

Density of air, ρ = 1.2 kg/m³

Volume of the balloon = 4/3πr³

Volume = 4/3 * 3.142 * 5.45³

Volume = 4/3 * 3.142 * 161.87

Volume = 4/3 * 508.76

Volume = 678.35 m³

Density = mass / volume

Mass = Density * volume

Mass = 1.2 * 678.35

Mass = 814.02 kg

Velocity = distance / time

Velocity = 106 / 21.5

Velocity = 4.93 m/s

If it starts from rest, 0 m/s, then the final velocity is 9.86 m/s

acceleration = velocity / time

acceleration = 9.86 / 17 m/s²

acceleration = 0.58 m/s²

The mass dropped from the balloon decreases Mb and increases buoyancy

F = ma

mg = (Mb - m) * a

9.8 * m = (814.02 - m) * 0.58

9.8m * (1/0.58) = 814.02 - m

16.89m = 814.02 - m

16.89m + m = 814.02

17.89m = 814.02

m = 814.02 / 17.89

m = 45.50 kg

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