Answer to Question #78470 in Physics for Sid

Question

A spherical helium balloon has mass 1.9 g and radius 9.71 cm. How fast will it be going 1.19 s after being released? Use 1.225 kg/m3 for the density of air. Ignore air resistance.

Answer 1

Balloon volume (it’s spherical):
V_b=4π/3 r_b^3=4π/3 (0.0971 m)^3=3.83∙〖10〗^(-3) m^3
The buoyant force is given by
F_b=ρ_a∙g∙V_b,
where g=9.8 m/s^2 – acceleration due to gravity.
Force of gravity:
F_g=m_b∙g
The resulting force acting on balloon is given by difference of these two forces:
F=F_b-F_g=g(ρ_a∙V_b-m_b )
The resulting acceleration:
a=F/m_b =(ρ_a∙V_b-m_b)/m_b g=((ρ_a∙V_b)/m_b -1)g=((1.225 kg/m^3 ∙3.83∙〖10〗^(-3) m^3)/(0.0019 kg)-1)9.8 m/s^2 =
=14.4 m/s^2
The speed of balloon after time t has passed is given by
v=a∙t=14.4 m/s^2 ∙1.19 s=17.1 m