Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A volcano erupts 50m below the sea level. A rock leaves the crater at 20 m/s at an angle 30 deg with the vertical line. The rock has a mass of 15kg. IGNORE WATER RESISTANCE. It gets out of the water, reaches a max height and falls back it.

When does it reach 30 below sea for the first and second times?

$a = <0, -9.8>$

$v = <20sin(30), -9.8t+20cos(30)> = <10, -9.8t+17.32>$

$s = <10t, -4.9t^2+17.32t>$

For the object to be 30m below sea level, y=20m, so $s_y=20 \implies -4.9t^2+17.32t-20=0$.

However, no real solutions exist to this. This is so simple, I apologize for asking, but what am I overlooking?

share|cite|improve this question
Where is the volume of the rock? – Pygmalion Apr 18 '12 at 20:44
What do you mean by "$s=\langle 10t,-4.9\,t^2+17.32\,t\rangle$"? It doesn't fit dimensionally. — EDIT ok I see none of this fits dimensionally, you seem to assume units with $[s]=[t]$. – leftaroundabout Apr 18 '12 at 20:44
@Pygmalion, it's not given. – Brandon Amos Apr 18 '12 at 20:47
up vote 6 down vote accepted

If there are no real solutions, that means that speed is too small to throw that rock so high!

So the problem was never solved properly by the one who created it.

IMHO, if you also have volume of the rock, you could calculate buoyancy and that would reduce gravitational acceleration and possibly rock could get so high.

share|cite|improve this answer
So my analysis is correct? My professor is going to collect our work on this problem next week, but I wanted to make sure I did this right before emailing him about this. – Brandon Amos Apr 18 '12 at 20:51
Right, and it can easily be seen by noting $|\mathbf{v}|^2/2<h\cdot g$. – leftaroundabout Apr 18 '12 at 20:52
You are correct. I think professor forgot the volume of the rock. This is the whole point of putting the problem in the sea... – Pygmalion Apr 18 '12 at 20:55
@leftaroundabout excellent point, +1 – Pygmalion Apr 18 '12 at 21:01
@user28554 Nice working with you, you up-vote and also accept answers. Not common around here. – Pygmalion Apr 18 '12 at 21:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.