Consider that a coin is dropped into a wishing well. You want to determine the depth of the well from the $T$ between releasing the coin and hearing it hit the bottom. Suppose $T=2.059 s$. What is the depth $h$ of the well?
- 20.77 m
- 19.60 m
- 23, 564 m
- 18.43 m
- 39.20 m
To find out the time it takes for a coin to drop to the bottom of the well, we know from equations of kinematics that $h=\frac{1}{2}gt^2$, where $g\approx 9.80 m/s^2$. Thus the time for the coin to drop to the bottom is $t_1=\sqrt{\frac{2h}{g}}$. The time for the sound to travel back up the well is $t_2=\frac{h}{v_{sound}}$. We want to solve for $h$, and we know that $t_1+t_2=T$ so we have \[\sqrt{\frac{2h}{g}}+\frac{h}{v_{sound}}=T.\] Using $T=2.059 s$, $g=9.8 m/s^2$, $v_{sound}=340 m/s$, we obtain $h=19.6 m$ and $h=25400 m$ upon solving the equation. Obviously $h=19.6 m$ is the correct solution while the other solution is an extraneous one that we got in the process of solving for $h$.
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There are a lot of questions on the GRE. You need to move quickly. Try to solve the problems without 'solving' them completely. Do rough calculations since you don't get a calculator. Usually a little physical intuition can lead to the correct answer in just a few seconds for each question.
For this one, instead of spending valuable time messing with the quadratic equation, instead solve it iteratively.
As you point out, the zero order approximation for the fall height is (roughly)
$ h= 1/2gt^2 \sim 1/2* 10*4 \sim 20m$
and you know that the true answer should be less than this given the fact that some time must be spent on sound propagation. So, within a couple seconds we go from 5 choices down to just 2: 18.43m and 19.60m with 20.77 perhaps being a possibility given that we estimated.
So, lets do a first order correction. Let's compute the time it takes for sound to go this distance.
$ dt=h_0/v_s \sim 20m/(340m/s) \sim 1/17s $
and lets adjust the height with the first order variation
$ dh=gt_0 dt \sim 10*2*1/17 \sim 1m$
So we suddenly know that we should be right around 19 m, so go with B.
Thanks for the alternate solution, that's quite clever!
A) the numerical quantities are far too close together for you to be able to reasonably solve these by hand, quickly. note the approximation above even doesn't work.
B) the answer choices on all ETS questions are sorted by ascending order.
what you are doing with your website is training people to think with the idiosyncrasies in your questions, which doesn't help.
I obtained the question from this book here. It seems sort of odd to be a GRE question but I'm quite sure it is.
As for "training people to think with the idiosyncrasies" in my questions: I'm not sure I know what you mean, so I offer no rebuttal.
Good day, sir.
You got me, but the method is sound, given that the answers are so close together, you really should spend a little more time on the computation to get another digit of accuracy.
Careful, those practice books make up their own questions. There have only been 4 official Physics GREs ever released. You can obtain them at the bottom of this page: http://www.physics.ohio-state.edu/undergrad/ugs_gre.php
I recommend you look at those, as quibblingtremors pointed out, the GRE questions typically are not so numerically picky, and studying from questions that are might get you in the mind set of working the problems out fully, which you do not have time to do no the real test.
Thanks for the great resource, I'll definitely be more picky where I choose my questions from now. Also, I think I'll emphasize speedy solutions and then give full solutions after. Your input has been really helpful, keep it coming!
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