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Hooray some mention of Japan in the syllabus!

Still in there. You probably hadn't done it at the time of the trial.

yep that's it.

oh of course :D Because it's horizontal, that's your horizontal velocity which doesn't change. So how do you get range? x velocity mutiplied by time. The time in this case is the time it takes for the projectile to fall 50m to the ocean. i.e. use r=ut+0.5(at^2) After that it's just...

191.7m easy :D

lol you could still hear the interference when it was on 105.7FM Just spent the whole time playing around with the induction coil and sparking thing and the radio... BTW does anyone know what the sparking thing is that you attach to the induction coil? It's not the same as Hertz', but does...

hehe it didn't matter...

http://www.boredofstudies.org/community/showthread.php?s=&threadid=17264 Exactly the same method.

This is hilarious...

And that's the practical?

Oh shit... not this practical. It's some crappy egg carton thing or with checker boards. It's so stupid I don't think I did it. Did anyone here do this practical? How did you do it?

Yes but Jacaranda is just excessive. Come on, they're training us to do the HSC, not to become actual Physicists.

Can someone tell me what the first hand investigation to demonstrate the motor effect is? I think I missed that one...

The Hertz Experiment isn't a first hand investigation. I'm talking about ones we're meant to have done by ourselves. What are the 3 you're thinking of?

Now time for some practical speculation :D I speculate: AC induction motor practical. What do you people think?

Used an induction coil to create sparking across some apparatus, then turned on a radio in the room and listened to interference. A 2 mark question at the most :P

Not really, because you're just subbing into formulas the same way as we're doing. You're just acknowledging that at the maximum height, y velocity is zero and using the formula r=ut+0.5(at^2) Note s = r = displacement. I think this syllabus actually refers to it as r, not s. If you...

You're the one who's making it too complicated by using a whole lot of steps.

Maybe someone copied out the question incorrectly...

Uh this is a very simple problem... Using v^2=u^2+2ar The answer is u=8m/s