The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and 9.0 kg, The terminal speed (in m/s) of Sphere 2 will now be:

Complete question is;

a. Two equal sized and shaped spheres are dropped from a tall building. Sphere 1 is hollow and has a mass of 1.0 kg. Sphere 2 is filled with lead and has a mass of 9.0 kg. If the terminal speed of Sphere 1 is 6.0 m/s, the terminal speed of Sphere 2 will be?

b. The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and 9.0 kg, The terminal speed (in m/s) of Sphere 2 will now be

A) V_t = 18 m/s

B) V_t = 10.39 m/s

Explanation:

Formula for terminal speed is given by;

V_t = √(2mg/(DρA))

Where;

m is mass

g is acceleration due to gravity

D is drag coefficient

ρ is density

A is Area of object

A) Now, for sphere 1,we have;

m = 1 kg

V_t = 6 m/s

g = 9.81 m/s²

Now, making D the subject, we have;

D = 2mg/((V_t)²ρA))

D = (2 × 1 × 9.81)/(6² × ρA)

D = 0.545/(ρA)

For sphere 2, we have mass = 9 kg

Thus;

V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρA))]

V_t = 18 m/s

B) We are told that The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1.

Thus;

Area of sphere 2 = 3A

Thus;

V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρ × 3A))]

V_t = 10.39 m/s

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