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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 ?

Mr. Y
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  • I think your doubt in the answer referred to is the assertion "When acceleration is constant, the instantaneous velocity at the midpoint of a time period is equal to the average velocity for [the] time period." That is incorrect, but at whatever point in a one second period the average velocity occurs, it will occur exactly one second later in the next such period, so the answer obtained was still correct ! – true blue anil Dec 28 '15 at 12:03

2 Answers2

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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,

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

  • why is it $2T+1$ shouldn't it be just $1$ ? – Mr. Y Dec 28 '15 at 08:59
  • There is a seeming anomaly in the equation as the terms are no homogeneous. This disappears if you write $\delta t$ instead of $1$, and the answer is $H+g,\delta t^2$. –  Dec 28 '15 at 09:00
  • @Mr.Y: please do the complete substitution by yourself. –  Dec 28 '15 at 09:01
  • I am not asking for that @Yves Daous.I am asking the concept behind that,because $2T+1$ seems to me as the $1$ second later the two $1$ seconds interval... – Mr. Y Dec 28 '15 at 09:02
  • @Mr.Y: understand my previous comment and perform that. –  Dec 28 '15 at 09:03
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    My apologies.Thanks for the clarification.Mathematical arguments are indisputable :) – Mr. Y Dec 28 '15 at 09:11
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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

true blue anil
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  • "so must the average velocity": you should prove that. –  Dec 28 '15 at 10:08
  • Well, I presumed that tackling such a question, OP would easily be able to figure out that in a straight line relationship, for any two points $1$ second apart, the difference in velocity would be the same. – true blue anil Dec 28 '15 at 10:23
  • The OP was stuck on the question, wasn't he ? –  Dec 28 '15 at 10:31
  • Ok, I am editing. – true blue anil Dec 28 '15 at 10:43
  • When you say "velocities at time $0,1,2$ " seconds do you mean velocities during those time intervals ?Or are we assuming that the bricks falls $40$ meters $1$ second later it was released from the building(quite unrealistic ) ? – Mr. Y Dec 28 '15 at 12:21
  • Not time intervals, at those points in time, with $t=0$ set at the start of the second in which it falls $40$ m, with initial velocity $u$ – true blue anil Dec 28 '15 at 12:30