Question

An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its speed, v, and obeys the equation Fdrag=(12.0N⋅s/m)v+(4.00N⋅s2/m2)v2. What is the terminal speed of this object?A. 72.2 m/sB. 12.6 m/sC. 47.3 m/sD. 34.2 m/sE. 6.45 m/s

Answer 1

The terminal speed of this object is 12.6 m/s

Step-by-step explanation:

It is given that,

Mass of the object, m = 80 kg

The magnitude of drag force is,

`F_(drag)=12v+4v^2`

The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.

`F_(drag)=mg`

`12v+4v^2=80* 9.8`

`4v^2+12v=784`

On solving the above quadratic equation, we get two values of v as :

v = 12.58 m/s

v = -15.58 m/s (not possible)

So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.