Question

A lighter than air balloon and its load of passengers and ballast are floating stationary about the earth. Ballast is weight of negligible volume that can be dropped overboard to make the balloon rise. The radius of the balloon is 6.25 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 105 m in 15.0s. (accelerated motion)

Answer 1

First Find the upward acceleration.

d = 105 m
a = ???
vi = 0
t = 15 sec

d = vi * t + 1/2 at^2
105 m = 1/2 a * 15²
a = 210 / (15)^2
a = 0.933 m/s^2


Now Find the Volume of the balloon and from that the mass.

V = 4/3 * pi * r^3
r = 6.25 m
V = 1022.65 m³.

density_of_air = m/V
1.29 = m / 1022.65
m = 1319.22 kg


Then the force keeping this in equilibrium =
F = m* g
F = 1319.22 * 9.81 = 12941.58 N


Find the mass needed to cause the upward acceleration.

12941.58 = (1319.22 - x) * (9.81 + 0.933)
12941.58 = (1319.22 - x) * 10.743
1204.65 = 1319.22 - x
x = 114.56 kg