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first of all, put any gaming consoles in the cupboard! if you can't study because you're thinking there's too much to do (e.g. doing a full three hour practice maths paper) then break down your work into smaller tasks (e.g. start by doing questions 1 and 2 for the last few hscs and check your...

i had the exact same problem, and i only fixed it by replacing the RAM

juai - 99.80 sam - 99.75 uai - 99.70

If initial velocity is 80m/s: u(h)=80cos45 u(v)=80sin45 range = u(h) x t(v) 150 = 80cos45 x t(v) t(v) = 2.65 s r= ut + 0.5at^2 = 80sin45 x 2.65 + .5 x -9.8 x 2.65^2 = 115.5 m hope that makes sense

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2006exams/index.html

17 for essay 12 for creative

I have no idea about question 1 either. I got an answer for question 2, although it involves simultaneous equations and a trigonemetric property i just learnt about in 3 unit maths (*), so there must be an easier way. (Let # be the angle we have to find) u = 30m/s u(horizontal) =...

You assumed that upwards was positive, since you said displacement = -10m and acceleration due to gravity is -9.8ms-2. But then wouldn't the downwards (i.e. vertical) velocity have to be negative as well? (i.e. -7.5ms-1) then -10= -7.5t-4.9t2 4.9t2+7.5t-10=0 Using the quadratic forumla...

isn't distance supposed to be 0.15m (=15cm) rather than 0.015m (=1.5cm)? r=ut+.5at^2 0.15 = 0 + .5 x -9.8 x t^2 t = 0.17s r=ut 250=u x 0.17 u=1428.87 m/s (i only started yr 12 as well, so i could be wrong too)