Physics word problem (challenge problem!)

To determine the height of a model airplane caught in a tree, an arrow is shot vertically upward next to it. The arrow passes the plane on the way up at .95 s after being shot upward, and at 1.60 s after being shot upward on the way back down. What is the height of the model airplane, and what was the initial speed of the arrow? (You may assume that the arrow is being shot from ground level). This is a bit of a challenge problem!

Time to final height = (1.6/2) + .95, = 1.75 secs.
Launch speed of arrow = (gt), = 9.8 x 1.75, = 17.15m/sec.
Height attained by arrow = (v^2/2g), = 15 metres.
Distance arrow drops in (1.6/2) = 0.8, secs = 1/2 (t^2 x g), = 3.14 metres.
Height of plane in tree = (15 - 3.14) = 11.86 metres.

This is too good of a challenge for me to show entire solution but I'll start it and reduce the challenge somewhat:

1.60 - 0.95 = 0.65 s = time arrow takes to rise from plane height to max height and fall back to plane height.

by symmetry: time for arrow to fall from max height to plane height = 0.65/2 = 0.325 s
t = √2h/g
(0.325)² = 2h/g
9.81(0.325)² = 2h
h = 4.905(0.325)² = 0.518 m = distance from arrow's max height to plane height
sH= 1/2gT² {where T = time for arrow to rise or fall to or from max height = H}
Vo = initial speed of arrow
Vo/g = time arrow takes to reach max height

can someone take it from here?