Distance covered by a brick in $1$ second time interval

Question

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 ?

Answer 1

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,

$$h(T+2)-h(T+1)=\frac{g(2T+3)}2=H+g.$$

Answer 2

More simply, the SUVAT equation $v = u + at$ shows that the "final" velocity increases by $a\; (= g)$ every second, thus so must the average velocity.(**)

Hence distance travelled in the next second $ = (40 +g)\cdot1 = 40+g$

(**) Explanation :

Velocities at time, $0,1,2$ seconds: $u, u+g, u+2g$
Average velocity $0-1: u + 0.5g = 40 =$ distance travelled in first second
Average velocity $1-2: u + 1.5g =40 +g =$ distance travelled in next second