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Dynamics

A = 0

For A=0 one is left with the equation of motion (differential equation) given by

dv/dt = -g + B×v²

However, here we must first pay attention to the direction of the friction force Bv², since v² is always positive.
When the object moves upward, then the friction force should be downward, hence B negative.
But, when the object moves downward, then the friction force should be upward, hence B positive.
Let us assume that the heighest position is reached at t=0. Then we end up with two different differential equations:

dv/dt = -g - |B|v2	for	t < 0	(upward)
dv/dt = -g + |B|v2	for	0 < t	(downward)

The upward equation is solved by (v_term and T are positive constants)

v = v(t) = -v_term×tan(t/T)   for   -πT/2 < t < 0

as one observes from

dv/dt = -(v_term/T)×{1+tan²(t/T)} = -(v_term/T) - (1/v_termT)×v²

by equating v_term=√g/|B| and T=1/√g|B| (see exercise 3c).

The downward equation is solved by

v = v(t) = -v_term×tanh(t/T)   for   0 < t

as one observes from

dv/dt = -(v_term/T)×{1-tanh²(t/T)} = -(v_term/T) + (1/v_termT)×v²

by once more equating v_term=√g/|B| and T=1/√g|B|.



the full solution