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Hahahahahhahahahahaah That's Classic!

Yeah I was asking because I know for loan repayments in 2U, you cannot get full marks if you just use the formula that the General Maths guys get given.

And I assume you got full marks?

Okay yeah, I get that that it's above the graph in this case, but why is that though? Just so I don't get confused with other questions like this.

Yeah but this is 2U or 3U stuff, so I dunno if I'd get the marks if I used that in a 2U/3U exam :P

Thanks.. I just have 1 question though, Is this area below or above the graph? If it's above why is that? Thanks

Here we go again... -_- I know how to do this, thank you very much. No I'm not. So when a ball hits the ground at a velocity, v=36m/s, its velocity changes to v=35, v=34, .. then v=-35 for an infinitesimal amount of time and then its velocity reaches v=-36m/s? If its velocity reaches zero (ie...

Hi, I'm pulling hairs with this question. Thanks if you can help me! Find the exact area bounded by the curve y=In x, the x-axis and the lines x=2 and x=4.

Read my post again, I said that the rate of change of velocity if CONSTANT. Therefore, the rate that the velocity changes from Vy=2 to Vy=1 is the same as the rate that the velocity changes from Vy=1 to Vy=0 and from Vy=0 to Vy=-1 and so on. You're saying that the projectile stops at Vy=0. If it...

In your case, it's collision, NOT constant acceleration (or rate of change of velocity), which makes it different. It goes through zero yes, but if the velocity was say, v=36m/s downwards initially, then according to you it should go through v=35, 34, 33, ... ,-36. It doesn't mean that it stops...

No, I haven't contradicted myself... According to you the projectile stops for an infinitesimal amount of time, which is undefined at the top of flight (ie. at Vy=0). At various points towards top of flight the projectile has a velocity of say, Vy=2 and then this changes to Vy=1 which changes...

If there is no air (like in vacuum) then there is no air resistance. That is going too far into Physics/Maths, which is not what I intended. But it is different to a ball accelerating due to gravity and then falling back. My point was that, the time that it touches the ground is undefined...

Then your definition of 'stop' is different to mine. When I say stop, I mean that the projectile has no velocity for a certain period of time. What you're trying to point out is something a little bit different. First of all, the projectile has a velocity (that isn't zero) before it hits the...

Trebla, as I have said MANY times. I said that the projectile in a flight GOES THROUGH Vy=0. Like any other velocity, it goes through Vy=1, Vy=2, Vy=3, etc etc etc... Not disagreeing with that. But I'm saying that the projectile DOES NOT stop at Vy=0. It is just like any those other quantities...

Okay, so you're telling me that if you go on a planet with a VERY VERY VERY VERY LITTLE gravitational force, and you throw a ball vertically for a very short distance then the velocity/time graph will not be a straight line going through the x-axis (ie where Vy=0), but it would have a part where...

It doesn't add points to your UAI, it just lowers the UAI requirement that you need to enter university!

You shouldn't rely on scaling for a good UAI... If you did well in those highly scaled subjects then you would've gotten a good UAI. Evidently you did not. What level Maths did you do? I hope you don't consider that General Maths is hard! You did all sciences? Well for starters, not all...

Is that aligned? Looks like more of a percentage to me?

Yep, eAccelerator or some other caching extentions can really reduce load.

I did work experience there, and I got to read that letter that the BoS sent due to a large number of non-serious attempts. I was working with the Managing Director =D Oh and they were warned last year (2007), not 2006. Not surprising, since it's the Daily Telecrap.