When is bunjee jumper's acceleration maximum This video talks about the point where the acceleration of bunjee jumper is maximum

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*Veritasium. "When Is A Bungee Jumper's Acceleration Max?", YouTube, May 25, 2011.

I approached it by simple problem of oscillation
$Y=A(1- \cos{\omega}t$).........(since both displacement and velocity are $0$ initially)
$$V=A\sin{\omega}t$$
$$a=A\cos{\omega}t$$
This gives a) as the right answer. I.e at the start of the motion but the answer is d). I haven't considered $g$ as it is constant throughout the journey. What is the error in my concept?
. Is there an effective ,legitimate way to accurately justify the answer d
 A: you have to solve this differential equation:
$$m\,\ddot z+k\,z=-m\,g\tag 1$$
the solution is
$$z(t)=\frac{m\,g}{k}\,\left(\cos \left( {\frac {\sqrt {k}t}{\sqrt {m}}} \right) -1\right)
\tag 2$$
with $z(t)$ and eq. (1) you obtain the acceleration:
$$ \ddot{z}=-g\cos \left( {\frac {\sqrt {k}t}{\sqrt {m}}} \right)\tag 3$$
eliminate from eq. (2) the time $t$ and substitute to eq. (3)
you obtain :
$$|\ddot z|=\frac km\,z+g$$
thus the max acceleration is when z is max at point d
edit
of course you obtain the same result if you solve eq. (1)
for $|\ddot z|$
A: Two points that might help you resolve your confusion:

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*The video is implicitly talking about the magnitude of the acceleration.  If the jumper was undergoing simple harmonic motion for the entire trip, then the magnitude of the acceleration at (d) would be equal to the magnitude of the acceleration at (a).


*The cable is slack between (a) and (b).  This means that the only force on these jumper between these two points is gravity;  he is not undergoing simple harmonic motion.
