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Help meh coconut guy
so far i drew the graph
idk how to get the maximum height
ermm... is the arrow released parallel to the ground
no its released from 1.39
and its a parabola so it won't be parallel to the ground
its curvy like a parabola ._.
ITS A PARABOLA
You can have a parabola with an initial angle too though!
According to your graph the max height would be 1.39m, however the question seems to easy
:O i just want to know how to find the maximum
Hmm let f(x) = ax^2+bc+c f(0) = 1.39 f(18) = (Height of point A) 16cm? f(45) = 0
c = y intercept c = initial height c= 1.39. You see where I'm going?
no.. solve the system of equations. plug in f(0), f(18) and f(45)
since we already know c... which is 1.39
nope I'm really lost
a(0)^2+b(0)+1.39=0 a(18)^2+b(18)+1.39=height of arrow a(45)^2+b(45)+1.39=0
I suppose the arrow's height should be 0.16m from the ground...
1.71 is wat im getting
I found this yahoo answers just now about the exact question. They use a similar approach. However I'm confused on how you would find the height of the arrow, the diagram only gives the height of the sections https://answers.yahoo.com/question/index?qid=20121015160405AAfjtCu
^ I'm not quite sure what the height of point A is... Most likely 0.16m from the ground since it lies 4 section above the bottom of the target
I am almost certain that the question does not give enough information to solve this as it stands. You NEED to know the height of point A from the GROUND and that is not given I guess we also have to assume that they mean "if the target was NOT present the arrow would land at 45m" ot htat eh second shot without the target is EXACTLY the aame trajectory. |dw:1435311494413:dw|
OK - THAT answer above has got more information - and gives the value for the height of A as required. it IS solvable once you have that information @Icedragon - was the height of the target given on YOUR drawing?