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
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?
2 answers
So for A is the answer just G??
For d is the answer -3/4mg
For d is the answer -3/4mg