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?

Answer 1

Answer:

 Terminal velocity of object = 12.58 m/s

Explanation:

 We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.

Gravitational force = mg = 80 * 9.8 = 784 N

Drag force = $12.0v+4.00v^2$

Equating both, we have

    $784=12.0v+4.00v^2\\ \\ v^2+3v-196=0\\ \\ (v-12.58)(v+15.58)=0$

  So v = 12.58 m/s or v = -15.58 m/s ( not possible)

 So terminal velocity of object = 12.58 m/s