A lighter-than-air spherical balloon and its load of passengers and ballast are

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 6.66 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 120 m in 13.0 s.

Explanation / Answer

since the balloon is stationary at first we know that the weight of the contents of the balloon is exactly equal to the weight of air that it displaces. V = 4/3 pi r^3 M = V*1.29kg/m^3 ---> if we multiply this by g this is the upward force on the balloon where M is the mass of the air ok so let's leave these for now and find the acceleration we need d = 1/2 at^2 ---> 120 = 1/2 (a)(13)^2 a = 240/13^2 so let's now make our equation g*[M - m] = netforce = a*m m is final mass (after stuff has been thrown off) M-m is the mass of whats thrown off gM - m(g + a) = 0 We know g, a, and M so solve for m and you need to throw off M-m

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