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So this is what I did..

\[y = V_{iy}t+(1/2)at^2=>-40=V_{iy}t-4.9t^2\]

yay Turing's here

give me a minute to finish breakfast plz

Ok lol.. i just started eating breakfast too actually hahah

first thing I gonna do when my hands are more free is draw it... I always draw it

>.> i got 8.27 degrees actually.

|dw:1348331215780:dw|

What was the equation that you differentiated?
-40=180Tan(theta) - 63.5(Sec(theta))^2 ?

I don't think differentiation is required

just write sec^2(theta) as 1+tan^2(theta).

ohhhhhhh

never am i going to trust wolf ever again..

good choice :)

-40 = 180Tan(theta) -63.5(Sec(theta))^2 and just solve for theta right?

yup.

thanks a lot guys