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 6.22 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 194 m in 13.0 s.

Answer 1

t = 13.0 s
d = 194 m

s = 1/2*a*t^2
194 = 1/2 * a * 13.0^2
a = 2.30 m/s^2

r = 6.22 m
Volume of ballon = 4/3 * π *r^3 = 4/3 * π * 6.22^3
Volume of ballon, V = 1008 m^3

Buoyant Force, Fb = p*g*V = 1.29 * 9.8 * 1008 N
Fb = 12743 N

Calculating mass of ballon,
12743 = m*g
m = 12743 / 9.8
m = 1300.3 kg

Let the mass of ballast that must be dropped be mb.

Fb - (1300.3 - mb) * g = (1300.3 - mb) * a
12743 - (1300.3 - mb) * 9.81 = (1300.3 - mb) * 2.30
mb = 248 kg

Weight of the ballast = 248 * 9.8 N
Weight of the ballast = 2430.4 N