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Dynamics

Motion at constant speed

After falling a while, the plastic ball suffers so much air resistance that it compensates the gravitational attraction such that its speed becomes constant, v_term. In that situation we may assume

t >> T   or   t/T >> 1

In that case we also may perform approximations for our functions.
For large x one has:

exp(-x) ≅ 0

Hence, it follows that for large x

tanh(x) = (exp(x)-exp(-x))/(exp(x)+exp(-x)) ≅ exp(x)/exp(x) = 1

Furthermore, for large x

cosh(x) = (exp(x)+exp(-x))/2 ≅ exp(x)/2

Hence, for large x

log(cosh(x)) ≅ log(exp(x)/2) ≅ x-log(2)

When we define y₀' = y₀-(v_term/T)log(2) and substitute all that in the above solution, we find for t>>T

y(t) ≅ y₀' - v_termt   , v(t) ≅ -v_term   and   a(t) ≅ 0

which describes the uniform motion of the plastic bal at constant speed, just as we had expected.

graphical representation