Reply
Tue 28 Feb, 2006 06:21 pm
I'm trying to prove that the range of a projectile is the same at angle x and angle (90-x). I came up with the following after a series of equations which I'm certain I did correctly.
cosx sinx = cosy siny
cosx cos(90-x) = cosy (90-y)
I'm a bit stuck now. I know cosx = cos(90-y), but I can't just assume that.
Any help would be appreciated.
vertically
0 = vsinx - gt
t= vsinx/g
horizontally
range= 2 vcosx . t
= 2v(sqd) sinx.cosx/g
so range = K sinx.cosx
but
sinx.cosx = cos(90-x) sin(90-x)
QED
Remember that x=90-y. That is a given. You got to cosxsinx=cosysiny. Now substitute in x. You also don't have to prove that sinx=cos(90-x). You can take that as given also.
Stumble It!
Tweet This
Bookmark on Delicious
Share on Facebook
Share on MySpace