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Hee, congrats Gangles.McGee :D And good luck IceOnFire :) Why is it a very tought competition for the girls though? What about the guys? :confused: As for myself, I'm simply tired. x.x Gee, I've lost count of how many times I've said that in the past...year...already. <.< *yawn* :p

Bleh. How come I don't remember doing this question? :confused: Meh, I'm getting senile :p But yes, mr_brightside covered it pretty well, and just another point: AM and FM have to be carried out on carrier waves of SOME wavelength. Therefore the relatively wide range of these waves on the...

Or Getteral09 might be in year 10 or lower and just trying to learn more maths than that taught at school :p

If you assign the upwards direction as the vertical positive direction, then the acceleration due to gravity would be negative. If, however, you take the positive direction as the direction towards earth, which would be downwards for a projectile motion over a flat ground, then the acceleration...

f3nr15, since the object was launched from the ground (and assuming that it landed on the ground :p), the point of landing is level with the point of launch. Therefore, the displacement in the vertical direction, ie. (delta)y, is zero :) And no twilight1412, that formula I mentioned is actually...

Doesn't feel like year 12...yet. :p I mean we haven't really started on *that* much stuff at school...hmm. This is actually how I felt at the beginning of year 11. Meh.

You school assessment marks mostly is VERY different from the HSC marks you would get. How did people at your school go in previous HSC exams?

lol, you answered your own question there, someone has done it before with ESL :p Theoretically what subjects you pick shouldn't determine your UAI, plus you're doing all Category A subjects anyway. And what's wrong with IPT? <.< :p

Oh God. Bumping up of another old thread. :p Meh, I might as well join :D Well, I make notes on the textbook...and memorise them :p

Or you could use the formula (delta)y = uyt + 0.5ayt2 for (a), where (delta)y = 0, ay = g = -9.8ms-2 and uy = u sin(theta) = 300(sin 30 degrees) = 150m/s, which works regardless of the point of launch :)

Using the formula (delta)y=uyt+0.5ayt2, Since you know that ay is simply gravity ie. ay=-9.8ms-2, and uy=u*sin(theta) where u=15m/s and theta=60 degrees, and (delta)y is just -3m as the ground is 3m below the balcony, you get: -3 = 15 sin(60 degrees)t - 0.5*9.8t2. Then solve for t, and take...

S1M0, was that an error or something that your HSC year is "2007"? o.0

Cram. :p Seriously though, just read through them and try to actually understand them. Feel the logics behind it, and it'll be much easier :)

Well, you know that for the standard parabola x2=4ay, the vertex would be at (0,0) and the focus at (0,a). Then just turn it about the origin 90 degrees at a time for the other three standard parabolas :D And all parabolas with horizontal or vertical axes of symmetry can be written in the form...

lol dw, at least you know you wouldn't do that next time :p

XD

Why can't you? o.0 But you of course you WOULD know something at least :D

The method is already given by vulgarfraction above, so what's the problem? Assuming the target is level with the point of launch, and that g = 9.8ms-2 as vulgarfraction did, and follow through her method, you get sin 2x = 0.294, so 2x is approx. 17 or 163 degrees (to the nearest degree)...

Eww. Calculus. <.< I think I like "elementary methods" better XP

Maybe you should look at the prelim content for General maths? See if you know it/can do it quickly. If so you shouldn't worry about catching up with the HSC course :D As for 2u, how much effort are you putting into it right now, and what marks are you getting? Think your effort matches up...