Solution incoming..Need to figure out what I'm doing wrong.

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Solution incoming..Need to figure out what I'm doing wrong.

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A driver needs to drive her car across a 180m long canyon. The side of the canyon opposite his starting point is 40m lower than the side that he's starting on. His initial speed is 50 m/s. At what smallest possible angle should a ramp be built so that he makes it across the canyon.
So this is what I did..
\[V_{iy}=V_{i}Sin\theta\] \[V_{ix}=V_{i}Cos\theta\] \[y=-40m\] \[x=180m\] \[V_i=50(m/s)\] \[\theta-?\]

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Other answers:

\[y = V_{iy}t+(1/2)at^2=>-40=V_{iy}t-4.9t^2\]
\[V_{iy}=V_iSin\theta=50Sin\theta\] \[-40=50(Sin\theta)t-4.9t^2\] \[x=V_{ix}t+(1/2)at^2=V_{ix}t=>180=50(Cos\theta)t=> t=18/(5Cos\theta)\]
After I plug in the value of t into: \[-40=50(Sin\theta)t-4.9t^2\] differentiate and solve for t, i get a crazy answer. Does anyone know why?
Gotta run to a store, will be back in about half an hour, would appreciate it if anyone could take a look at this and tell me what's wrong with my solution.
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

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