Question

A 2.0g latex balloon is filled with 2.0g of helium. When filled, the balloon is a 28 cm diameter sphere. When released, the balloon accelerates upward until it reaches a terminal speed. What is this speed? Assume air density of 1.2 kg/m^3 and a drag coefficient for sphere of 0.47

Answer 1

Density of Helium is 0.18 kg/m³
Density of air is1.2 kg/m³
Difference is 1 kg/m³
volume of balloon is V

= ⁴/₃πr³

= ⁴/₃π(0.31/2)³

= 0.101 m³

upward force is 1 kg/m³ x 9.8 N/kg x 0.101 m³

= 0.99 N
subtract weight of 0.003 x 9.8

= 0.0294 to get 0.96 N

I used density of helium at STP instead of your weight number. They should be about the same. 0.101 m³ / 0.18 kg/m³ = 0.0182 kg or 18g but they are not...

but the 2.5g indicates no lift.

F = ma
a = F/m = 0.99/0.003 = 330 m/s²

Vt = √(2mg/ρACd)
Vt = √(2(0.003)(330)/(1.2)(π0.31²/4)(0.47))

Vt = 6.71 m/s


Vt = √(2mg/ρACd)
Vt = terminal velocity,
m = mass of the falling object,
g = gravitational acceleration,
Cd = drag coefficient,
ρ = density of the fluid/gas
A = projected area of the object.

Thanks and All the best