# Homework Help: Projectile:body falling down well

1. Jan 23, 2013

### CAF123

1. The problem statement, all variables and given/known data
A body of some mass $m$ is thrown up from the ground with a speed of $v_o$. Suppose that upon returning to ground level, the body enters into a deep well. What is the relation between the height of the well and the time it takes until the body hits the ground?

2. Relevant equations
$\ddot{x} = -g$

3. The attempt at a solution
I am a little confused with the wording of the question. Does that mean the height as a function of time or time function of height?

I assumed the former: On entrance to the well, by symmetry, $\dot{x}(0) = -v_o, x(0) = 0$. I have two scenarios which I think justify why $\dot{x}(t_G) = 0$. 1)If bottom of well soft, body will just stick. 2)If bottom of well hard, body will rebound to some height but this means it's velocity has to change direction so at some instantaneous moment (which I take just before it rebounds) it's velocity is zero. So $t_G$ = time to ground = $v_o/g$.
Let h be the height of the well. Then $-h = -v_ot_G - \frac{1}{2}gt_G^2$. Subbing in $t_G$ gives $h = \frac{3v_o^2}{2g}$, which doesn't make sense since the sign is wrong. I realise what I have here is an implicit function of time. Is this what the question wanted or is there something else?

2. Jan 23, 2013

### voko

You need to find the time it takes to reach to bottom of the well. You have correctly deduces that the return speed is equal to the initial speed (not by symmetry, but conservation of energy), and the direction of velocity is negative. Assume the depth of the well is -h, and find out the time. That is what the problem is about.

3. Jan 23, 2013

### CAF123

I see. So it is simply :$$t_G = \frac{-v_o \pm \sqrt{v_o^2 + 2gh}}{g}.$$
Forgetting what the question wanted, is there anything wrong with my analysis my opening post?

4. Jan 23, 2013

### voko

The velocity can become zero only when the body reaches its max height (again). Or, assuming some energy is lost due to resistance and/or collision with the bottom of the well, anywhere else. But I can' see why that is relevant in any case.

Re the formula you derived, you still need to select ONE solution.

5. Jan 23, 2013

### CAF123

It must be zero at some point though because the velocity is pointing downwards intially but when it hits the ground and rebounds it then points upwards, so there has to be a time when it is momentarily zero, right?

Perhaps I should use the soln with the -√. This will give the first time of contact with the floor and the later soln '+√' corresponds to that time when it rebounds and hits the floor again.

6. Jan 23, 2013

### voko

Yes, but what does that have to do with hitting the ground?

The formula you used initially - uniformly accelerated motion - does not have any provision for "rebounding". So that can't be it. Selecting the minus sign before the radical will result in a negative time. What does that mean, physically?

7. Jan 23, 2013

### CAF123

I suppose that is where I was going wrong - I was assuming when the body hit the ground it had zero velocity which led to my incorrect equation for h.

Is it if the ball went in the upwards direction? (which is unphysical)

8. Jan 23, 2013

### voko

Remember, at zero time, the body is at the top of the well, falling down. What does negative time really mean?

9. Jan 23, 2013

### CAF123

Would it be at some point when it is falling back down to earth?(falling back down to the ground where it enters the well)

10. Jan 23, 2013

### voko

It was definitely before it fell down to "ground zero". But it was also what the depth was -h. So when could that be?

11. Jan 23, 2013

### CAF123

Does it correspond to the same height as the well, going upwards? (Or to be more clearer, the neg. time is the time when the body reaches a height h above the ground?

12. Jan 23, 2013

### voko

The negative time is a solution when the body as at the bottom of the well.

13. Jan 23, 2013

### CAF123

Oh, I thought this was the positive time. So what then is the postive time? Your words 'it was definitely before it fell down to 'ground zero' indicated to me that the negative time occured before the body struck the ground, (assuming ground zero is the place where it's velocity is $-v_o\hat{j}$)

14. Jan 23, 2013

### voko

Both solutions correspond to the position of the body at the bottom of the well. But one was before (negative time) it was at the top of the well, the other (positive time) is after that time. So which one is the solution of the problem?

Bonus question: what does the other one mean?

15. Jan 24, 2013

### CAF123

So I would say the positive radical was the answer to the problem.
It is not very clear to me what the negative time would represent here. I don't see another time at which the body would be at -h. Thanks

16. Jan 28, 2013

### CAF123

The negative time would correspond to that time when the particle reached -h if it were dropped at some time in the past? Is this right and if so since this is unphysical, it is neglected.

17. Jan 28, 2013

### voko

Remember, the trajectory consists of two parts: one going up, and one going down. The negative time corresponds to the going-up part. If the body was launched from the bottom of the well at the negative time you have found, then it would hit the bottom at the positive time.