A sphere is dropped from a great height and when its speed reaches 20 m/s,its acceleration downward is 8.4 m/s².If the drag force is proportional to the square of the speed,what will be its acceleration downward when its speed reached 40 m/s? A) 2.6 m/s² B) 4.2 m/s² C) 8.0 m/s² D) 0 m/s² because it has now reached terminal velocity.

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Quizplus · Accepted Answer

We can use the formula of drag force, which is given by Fd = -kv^2, where Fd is the drag force, v is the speed, and k is a constant of proportionality. The negative sign indicates that the force is opposite to the direction of motion. Now, we can use Newton's second law of motion, which states that the net force on an object is equal to its mass times its acceleration (F = ma). When the sphere is dropped, the only force acting on it is its weight, which is given by Fg = mg, where m is the mass of the sphere and g is the acceleration due to gravity. Thus, we have Fg - Fd = ma, or mg - kv^2 = ma. We can rearrange this equation to solve for the acceleration, which is a = g - (k/m)v^2. When the speed is 20 m/s, we know that the acceleration is 8.4 m/s^2. We can use this information to find the value of k/m, which is (g - a)/v^2 = (9.8 - 8.4)/(20^2) = 0.0004 m^-1. Now, we can use this value of k/m to find the acceleration when the speed is 40 m/s: a = g - (k/m)v^2 = 9.8 - (0.0004)(40^2) = 4.2 m/s^2. Therefore, the answer is B.

A Sphere Is Dropped from a Great Height and When