When the plastic ball is in its highest position it has no velocity.
Near the highest point its velocity is small.
Hence, we expect that then also the air resistance is negligible,
in which case the plastic ball behaves as a freely falling object.
Here, we will study this for the solution which applies to
t>0, i.e.
y(t) = y0 - vtermT×log(cosh(t/T))
for
0 < t
v(t) = -vterm×tanh(t/T)
for
0 < t
a(t) = -(vterm/T)
+ (1/vtermT)×v2(t)
for
0 < t
This solution describes the motion of the plastic ball
after it reached the highest position and returns back to the ground.
Near the highest position of the plastic ball,
we have that t is small,
just a little bit larger than t=0.
So, we may assume that
t << T or
t/T << 1
In that case we may perform approximations for our functions.
For small x one has:
exp(x) ≅ 1+x and
exp(-x) ≅ 1-x
Hence, it follows that for small x
tanh(x) = (exp(x)-exp(-x))/(exp(x)+exp(-x)) ≅ x
For the cosh we need one term more
in the expansions of the exp functions:
exp(x) ≅ 1+x+x2/2
and
exp(-x) ≅ 1-x+x2/2
Hence, it follows that for small x
cosh(x) = (exp(x)+exp(-x))/2 ≅ 1+x2/2
and, furthermore
log(cosh(x)) ≅ log(1+x2/2)
≅ x2/2
When we substitute all that in the above solution, we find