A box with a mass of 0.6-kg falls from a very tall building. While free-falling it is subjected to a frictional drag force given by Fdrag = bv2, where v is the speed of the object and b=5.5 N•s2/m2 What is the terminal speed that the box will reach?

Answer:1.03m/sExplanation:According to the questionFdrag = bv²now Fdrag = weight of the boxthat is Fdrag = mgwhere m = mass of the box, 0.6 Kgand g = acceleration due to gravitythus,Fdrag = bv²   becomesmg = bv²making v the subject of formula[tex]v = \sqrt{\frac{mg}{b} }[/tex][tex]v = \sqrt{\frac{0.6 * 9.8}{5.5} }[/tex]v = 1.03m/s

Answer:the terminal speed that the box will reach v = 1.0355 m/sExplanation:The drag force is the resultant force component that acts in the relative motion direction of the body.Here we have Fdrag = bv²But Fdrag = mgTherefore F = 0.6 × 9.81 =5.886 Nv² = Fdrag / b = 5.886 N/(5.5 N s²/m²)v = 1.0355 m/s