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my other way i did it was an algebra bash, ended up with 200 but took 3-4 pages :(

oik here goes... notice similar triangles... x/10 = (49 - 4.9t^2) / 4.9t^2 find dx / dt = [(4.9t^2)(-98t) - (490 - 49t^2)(9.8t)] / (4.9t^2) sub in t = 1 and dx/dt = -200 ie v = 200m/s to the left

haha i think i got it out, can u pm or post an answer (cbf typing 2pages out for nothing) would like to see CMs answer for this, like the other qn tho :P

the 2k.pi is just the general solution

have u got answers? heres my attempt AC = AE + EC (3) AE / DC = AB / DB (from parts i and ii) AE = (AB.DC / DB) (1) EB / EC = AD / AB (couldnt find a better way of doing it :( so i proved those are similar) AD.EB = AB.EC EC = (AD.EB) / AB (2) sub 1 and 2 into 3 AC = (AB.DC /...

this isnt really a thinking question, derive the forumulas, sub values in and ur set !

cause arctan 1 = (pi) n + pi/4

ahh yes..but if they give a domain

isnt it tan(x+y) = 1 x+y = pi/4 which is a straight line?

can you clarify this ? 49-4.9t2 t^2 ? and im assuming 4.9 not 4x9

ahh yes :) nice CM thats alot better than my unelegant 1 1/2 page algebra bash :(

:o how do you hide answers? anyway i only posted it cause i thought u neded help with it

:( i used induction

Let angle (BCA) = @ (BOA) = 2@ (ABX) = 2@ (where x is on cb subtended) .:. (BCA) + (CAB) = 2@ (exterori angle of a tri = sum of opp interor angles) @ + (CAB) = 2@ (CAB) = @ (BCA) = @ .:. ABC iso

hehe let t = tanx/2 lhs = (1 + t) / (1 - t) rhs = (1 + t^2 + 2t)/(1 - t^2) = (1 + t)^2 / (1+t)(1-t) = (1 + t) / (1 - t) = lhs

i found when @ = 45 (ie max distance) then let the point p(x,y,z) be a point representing the water path. i found that p(x,y,z) is equidistant from the line (0,31c/8,z) and the point (0, -c/8, 0) hence the pt P is a locus representing a parabola..etc the reason i said find the co-ords of z...

"argh, your not in year 12, are you? :-\" err yes allright then if you dont want the answer for ii) ill give some hints for those wanting to try it. getting (15c)/8 shouldnt be that hard - just alot of algebra then consider the height of the wall the y axis the 'jet' on the x axis and...

ahh ii) took me lik 2 pages, ill scan and post it 2moorow

(i) Prove that the jet can reach the wall above ground level if and only if V >(gc)1/2. (ii) If V = 2.(gc)1/2 prove that the portion of the wall that can be reached by the jet is a parabolic segment of height (15c)/8 and area (5.sqrt15(c^2))/2. i)) i cant be bothered deriving formulas...

so what did you get for Q and R ?? (cant access it :( )