Distance covered by a brick in $1$ second time interval 
A brick is dropped from the roof of a tall building.After it has been
  falling for a few seconds ,it falls $40.0$ meters in a $1.00$-s time
  interval.What distance will it fall during the next $1.00$ seconds?
  Ignore air resistance.

I've seen on yahoo answer that the solution provided for second part of the question,namely the distance covered in the second interval of $1.00$ second,can be found just by realizing that the midpoint of this interval occurs $1$ second later the midpoint of the first interval(so this would allow us to calculate instantaneous velocity).
However I am a bit skeptical ,because after the midpoint of the first interval the brick's velocity kept raising in magnitude,so I would think that  it hasn't necessarily have to be at the midpoint of the second interval exactly $1$ second later after the first one.
This way I would be assuming that the brick's velocity is constant over that interval,while it isn't.
Another point of confusion is that I can find many intervals of $1$ second where the brick falled a distance of $40$ meters..
Question
Can you guys make this clear ?Am I right or wrong ?If I am right how would I solve the problem ?

By the well-known relation, the space traversed by the brick follows
$$h(t)=\frac{gt^2}2.$$
You are given that
$$h(T+1)-h(T)=H=\frac{g(2T+1)}2.$$
Then,
$$h(T+2)-h(T+1)=\frac{g(2T+3)}2=H+g.$$
