1. A sphere with a mass of 10 kg and radius of 0.5 m moves in free fall at sea level (where the air density is 1.22 kg/m3). If the object has a drag coefficient of 0.8, what is the object’s terminal velocity? What is the terminal velocity at an altitude of 5,000 m, where the air density is 0.736 kg/m3? Show all calculations in your answer.

The value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.

Given data:

The mass of sphere is, m = 10 kg.

The radius of sphere is, r = 0.5 m.

The density of air is, .

The drag coefficient of object is, .

The altitude is, h = 5000 m.

The density of air at altitude is, .

The mathematical expression for the terminal velocity of an object is,

here,

g is the gravitational acceleration.

A is the area of sphere.

Solving as,

Now, the terminal velocity at the altitude of 5000 m is given as,

Solving as,

Thus, we can conclude that the value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.

Learn more about the terminal velocity here:

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The terminal velocity at sea level is 7.99 m/s

The terminal velocity at an altitude of 5000 m is 10.298 m/s

Explanation:

mass of sphere m  = 10 kg

radius of sphere r = 0.5 m

air density at sea level p = 1.22 kg/m^3

drag coefficient Cd = 0.8

terminal velocity = ?

Area of the sphere A = = 4 x 3.142 x = 3.142 m^2

terminal velocity is gotten from the relationship

where g = acceleration due to gravity = 9.81 m/s^2

imputing values into the equation

= 7.99 m/s

If  at an altitude of  5000 m where air density = 0.736 kg/m^3, then we replace value of air density in the relationship as 0.736 kg/m^3

= 10.298 m/s

true

explanation:

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s