A body of mass m falls from rest through the air. A drag force D = bv^2 opposes the the motion of the body. The acceleration due to gravity is g.

A) What is the initial downward acceleration of the body?

B) After some time the speed of the body approaches a constant value. What is the terminal speed vT?

I got this one to be sqroot(mg/b)

C) What is the downward acceleration of the body when v = vT/2?

A: since the initial v=0, only g is acting on the mass.

B: ok

C: D = bv^2 = b(√(mg/b)/2)^2 = mg/4
SO, net a = mg/4 - mg = -3/4 mg

So for A is the answer just G??

For d is the answer -3/4mg

To answer these questions, we need to use the equation of motion for the falling body, taking into account the drag force opposing its motion.

A) The initial downward acceleration of the body can be determined from the net force acting on it. The net force is the difference between the gravitational force and the drag force:

Net force = mg - D

Since the drag force is given as D = bv^2, we can substitute it in:

Net force = mg - bv^2

Using Newton's second law (F = ma) and rearranging the equation, we get:

ma = mg - bv^2

Dividing both sides by the mass (m), we obtain:

a = g - (bv^2) / m

Therefore, the initial downward acceleration of the body is (g - (bv^2) / m).

B) The terminal speed (vT) is the maximum speed that the body can attain when the drag force and gravitational force are balanced. At terminal speed, the net force on the body is zero:

Net force = 0

mg - bvT^2 = 0

Solving for vT, we get:

bvT^2 = mg

vT^2 = mg/b

Taking the square root of both sides, we find that the terminal speed, vT, is equal to the square root of (mg/b).

C) To find the downward acceleration of the body when v = vT/2, we can substitute this value into the equation for acceleration that we derived in part A:

a = g - (bv^2) / m

Substituting v = vT/2, we have:

a = g - (b(vT/2)^2) / m

Simplifying the equation, we get:

a = g - (b(vT^2)/4) / m

Since we know that vT^2 = mg/b (from part B), we can substitute this value in:

a = g - ((b(mg/b))/4) / m

Simplifying further, we find:

a = g - (g/4)

a = 3g/4

Therefore, the downward acceleration of the body when v = vT/2 is (3g/4).