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|dw:1397053258940:dw|

\[y=-\frac{ 1 }{ 2 }g t ^{2}+ v _{0} t+ y _{0}\]

that one ?

but i don't know about the intial velocity

i mean, they did not give that value

Yes, you have to find it. Hold on for one second. I'll attach one thing.

ok

\[|v _{x}|= v _{x0} \cos 40°\]
\[|v _{y}|= v _{y0}\sin40°\]

\[v =\sqrt{v ^{2} _{x}}+v ^{2}_{y}\]

if i use \[\tan 40°\] that would be the initial velocity ?

Do you have the answer key?

im going to check that . . .

it seems like it doesn't have the key answer

i know them but i have to solve this quickly

thanks anyway

please anyone tell me how should aproach this problem

so its just me and u huh? :D

lol please help! friend

i guess in the horizontal component there is not accelaration

correct!

so in certain way i was right about the tan 40°

the thing is i don't know where that comes from, i mean the equation

is there a way that not need derivation

http://www.triton.edu/uploadedFiles/Content/Resources_and_Services/Services/Academic_Success_Center/About_Us/projectiles.pdf
You do not need derivation for this, if you are allowed to use the equation the directly. The non-calculus proof is linked above ^ .
I can tell you WHY we need the equation. As we have an incomplete parabola, due to the height difference, a @Mashy has drawn above, it is necessary to use the parabola equation because the velocity then, can be calculated at that point. :)
This is the equation.
This equation is the equation used to get all the other equations of height and range. :)

And there in the equation, y is actually \(\delta y\) giving the change in height. :D

|dw:1397056685213:dw|

well im going to try in the first way . ..

just do the way i told u.. u ll get it :-/

|dw:1397057202246:dw|

well in te way you told me i have to multiple by G on the VERTICAL COMPONENT ?

i have a hard time understanding ur 'drawings' :P
jsut tell me what answer u get :P

lol

\[u = \frac{ t }{ -g*\sin 40*3.05 }\]

is that the equation for the vertical component ?

http://www.twiddla.com/1561200

i guess im wrong far so far fromget to the answer

\[\frac{ 10\frac{ m }{ s ^{2} }(10m)^{2} }{ (1.05m-10mtan40°)2 \cos40° } = v\]

no way i couldn't do it

@mashy what did i wrong dude ... xD