1. Jul 29, 2009

### SusanC1105

bsbtstrrbt

Last edited: Jul 30, 2009
2. Jul 29, 2009

### diazona

What have you done so far? If you don't show us any work you've been able to do, we can't tell you what you might be misunderstanding.

3. Jul 29, 2009

### SusanC1105

erererger

Last edited: Jul 30, 2009
4. Jul 29, 2009

### RoyalCat

That is incorrect.

The velocity of the rock on the way up at height $$x$$ is the same as the velocity of the rock on the way down at height $$x$$
You can prove this using kinematics ($$v_f^2=v_0^2+2ad$$), or using preservation of energy. But that is irrelevant for this problem, as it deals with just the upwards motion.

The first segment you're asked about is from 0 meters above the ground, to 5 meters above the ground. What is the rock's initial velocity (You can calculate this using the two approaches I cited above) and what is its acceleration? You can easily construct an expression describing the time it takes for it to traverse the 5 meters.

Now consider the second segment, which is from 15 meters above the ground, to 20 meters above the ground. What is the rock's velocity at the start of this motion and what is its acceleration? As with the previous segment, you can construct an expression for how long it takes to traverse the 5 meters, or any distance, really.

I suggest you solve this parametrically, as any exam questions are very likely to have follow-up questions which you will find your parametric solution very useful for solving.

EDIT: The answer in my spoiler is irrelevant. I misread the question. Your reasoning is still incorrect, though.

The velocity of the stone is not the same the way up as it is the way down. The acceleration is constant, that much is true, but the last 5 seconds of its flight are the 5 seconds before it hits the floor (20 meters below the throwing point) and first 5 seconds of its flight are when its first cast upwards. They are VERY different.
The formula that has all the answers in this case is $$v_f^2=v_0^2+2ad$$
Since we know the velocity is symmetrical with respect to the maximum height (The velocity of the stone 5 seconds before reaching the maximum height, is the same as its velocity 5 seconds after reaching the maximum height. The same can be said about its height), we'll make the argument as though our stone started from the maximum height, some $$x$$ meters above the throwing point, which is 20 meters above the ground.

Going by the above formula, when is the stone's velocity greater, in the last 5 seconds of its flight (When it clears the ADDITIONAL 20 meters of the height of the throwing point), or in the first 5 seconds of its flight?

Last edited: Jul 29, 2009
5. Jul 29, 2009

### SusanC1105

yjttytytjty

Last edited: Jul 30, 2009
6. Jul 29, 2009

### Amar.alchemy

I think its better to explain it in terms of "the direction of acceleration"

7. Jul 29, 2009

### RoyalCat

A bit busy now, will reply to the rest later.
Replies in bold.

8. Jul 29, 2009

### SusanC1105

dtyjtjdtj

Last edited: Jul 30, 2009
9. Jul 29, 2009

### shahin1234

Did you ever fugure out number 4,7,10, 12, 15 or 16, i got the rest if you need any?

10. Jul 29, 2009

### SusanC1105

thrhrhrth

Last edited: Jul 30, 2009
11. Jul 29, 2009

### shahin1234

Is there any that you need

12. Jul 29, 2009

### SusanC1105

srthsrthh

Last edited: Jul 30, 2009
13. Jul 29, 2009

### shahin1234

for 5 i used the PE=mgh and explained that height is the only thing changing. and KE= .5mv^2, the other ones im kinda iffy about. 4,10,15 are the worst.

14. Jul 29, 2009

### queenofbabes

Attempting to pick up where RoyalCat left off. Answers in bold.

Seems like you're still confused over 4. Once the string breaks, what are the forces acting on it? Apply Newton's First Law. That's all there is to it.

Last edited: Jul 29, 2009
15. Jul 29, 2009

### SusanC1105

rsthsrth

Last edited: Jul 30, 2009
16. Jul 29, 2009

### SusanC1105

srthhrs

Last edited: Jul 30, 2009
17. Jul 29, 2009

### queenofbabes

Well, what are your thoughts on the responses I've provided?

18. Jul 30, 2009

### RoyalCat

Replies in bold.