Question

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 5.80 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 199 m in 24.0 s.

Answer 1

Find the upward acceleration.

d = 199 m
a = ???
vi = 0
t = 24 sec

d = vi * t + 1/2 at^2
199 = 1/2 a * 24^2
a = 398 / (24)^2
a = 0.70 m/s^2

Step Two
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Find the Volume of the balloon and from that the mass.

V = 4/3 * pi * r^3
r = 5.80 m

V= 816.89 m^3.

density_of_air = m/V
1.29 = m/816.89
m = 1054 kg

Step Three
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The force keeping this in equilibrium =
F = m* g
F = 1054* 9.81 = 10340 N

Step Four
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Find the mass needed to cause the upward acceleration.

10340 = (1054 - x) * (9.81 + 0.70)
10340 = (1054 - x)* 10.51 Divide both sides by 10.51
985 = 1054 - x
x = 69 kg