at sea level 3.2 diameter is spherical balloon Travels at 22 metre per second at 3 kilometre altitude what speed will give the same drag force let CD be the same at both altitude

24.89m/s

Explanation:

We can calculate the drag force F to be:

F = 0.5 x CD x ρ x A x v²

When at sea level we have

h = 0

ρ1 = density of air = 1.225m³/kg

h = 3000 at altitude

ρ2 = 0.957m³/kg this is the density of air. We got this from the standard table of density variation

Because F1 = F2

0.5CD x ρ1 x A x V1² = 0.5CD x ρ2 x A x v2²

When we solve this out

p1v1² = p2v2²

V2² = p1v1²/p2

V2² = 1.225x 22²/0.957

V2² = 1.225x484/0.957

= 619.5

V2 = √619.5

= 24.89m/s

This is the speed

i mean i gues

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