A physics problem..help?

Question:An elevator without a ceiling is ascending with a constant speed of 10 m/s. A boy on the elevator throws a ball directly upward, from a height of 2.0 m above the elevator floor, just as the elevator floor is 31 m above the ground. The initial speed of the ball with respect to the elevator is 17 m/s.
1.What is the maximum height attained by the ball?
2.How long does it take for the ball to return to the elevator floor?


show work if possible..

Answers:
The ball is projected from a point (31 + 2)m above the ground, with an initial velocity relative to the ground of (10 + 17)m/sec.

1.
If h is the maximum height from where the ball is released, then:
0^2 = (10 + 17)^2 - 2gh
2gh = 27^2
h = 27^2 / 2g
Using g = 9.81 m/sec^2:
h = 27^2 / (2*9.81)
= 37.16m.
The ball was 33m above ground when released, and so the maximum height above ground is 33 + 37.16m = 70.16m.

2.
It's unclear where the time measurement is to start. I am assuming it is from the moment the ball starts descending.

The time t from the moment the ball is thrown up at 27m/sec until it starts descending is given by:
0 = 27 - gt
t = 27 / g = 27/9.81 = 2.75sec.
In that time, the lift floor has ascended 10*2.75 = 27.5m from its initial height of 31m.
It is therefore 58.5m above ground.

At time t after the ball starts descending, its height s above ground is:
s = 70.16 - gt^2 / 2

The height s1 of the lift floor above ground at the same time is:
s1 = 58.5 + 10t

The required time t is therefore when:
s = s1
70.16 - gt^2 / 2 = 58.5 + 10t
gt^2 / 2 + 10t - 11.66 = 0
4.905t^2 + 10t - 11.66 = 0
t = ( -10 +/- sqrt(100 + 4 * 11.66 * 4.905) ) / 2
= ( - 10 +/- 18.13 ) / 2.

Discarding the negative result:
t = 8.13 / 2 = 4.07 sec.

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