bahrom7893 Solution incoming..Need to figure out what I'm doing wrong. one year ago one year ago

1. bahrom7893

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.

2. bahrom7893

So this is what I did..

3. bahrom7893

$V_{iy}=V_{i}Sin\theta$ $V_{ix}=V_{i}Cos\theta$ $y=-40m$ $x=180m$ $V_i=50(m/s)$ $\theta-?$

4. bahrom7893

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

5. bahrom7893

$V_{iy}=V_iSin\theta=50Sin\theta$ $-40=50(Sin\theta)t-4.9t^2$ $x=V_{ix}t+(1/2)at^2=V_{ix}t=>180=50(Cos\theta)t=> t=18/(5Cos\theta)$

6. bahrom7893

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?

7. bahrom7893

@TuringTest @anemonix

8. bahrom7893

@AccessDenied @satellite73 @amistre64

9. bahrom7893

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.

10. bahrom7893

yay Turing's here

11. TuringTest

give me a minute to finish breakfast plz

12. bahrom7893

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

13. TuringTest

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

14. rajathsbhat

>.> i got 8.27 degrees actually.

15. bahrom7893

|dw:1348331215780:dw|

16. bahrom7893

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

17. TuringTest

I don't think differentiation is required

18. rajathsbhat

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

19. bahrom7893

ohhhhhhh

20. bahrom7893

never am i going to trust wolf ever again..

21. TuringTest

good choice :)

22. bahrom7893

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

23. rajathsbhat

yup.

24. bahrom7893

thanks a lot guys