A ball is thrown vertically upwards at speed v0. Assume drag force is proportional to v2.
a) Show that, while moving upwards, Newton's Second Law gives a = -g(1+v2/vt2) where vt is the terminal speed.
b) Take v0 = 3vt and solve for v(t), the time tmax at which it reaches max height and the maximum height it reaches. Express your results in terms of g and vt
a = -g(1+v2/vt2)
v = -vt tanh (t/T), where T = vt/g
y = -vt2/g * ln [cosh (gt/vt)]
The Attempt at a Solution
I got part A of the question simple enough, I'm having trouble with part B. I got the expression for v(t) which was the integral of the first equation. I'm having trouble looking for tmax and the max height. It sounds simple enough but I can't grasp around the idea on how to do it.