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Homework Help: Solve this linear equation

  1. Oct 9, 2009 #1
    1. I thought I was doing it right, but Quest said I have to wrong answer. Can some one solve it in steps and produce a final answer so I can compare?

    Tan 58=(vi-ayt)/axt

    where vi=5.6

    solve for t.

    I got 0.87 for t.

    Thank you.
  2. jcsd
  3. Oct 9, 2009 #2
    Could you please include your attempt to a solution?
  4. Oct 9, 2009 #3
    vi-at=tan 58 (at)
  5. Oct 9, 2009 #4
    Take a look at the second to last step. You have:

    6.44*t = -5.6

    Your answer should be negative.
  6. Oct 9, 2009 #5
    How do we have negative time though? (its the math to a physics problem)
  7. Oct 9, 2009 #6
    Does your problem require you to be in degrees or radians?
  8. Oct 9, 2009 #7
    Degrees, and my calculator was set on degrees.
  9. Oct 9, 2009 #8
    Maybe you should post the original physics problem. Is it possible you made an error setting the problem up :eek:?
  10. Oct 9, 2009 #9
    Initially (at time t = 0) a particle is moving vertically at 5.6 m/s and and horizontally at 0 m/s. The particle accelerates horizontally at 2.1 m/s2 . The acceleration of gravity is 9.8 m/s2 . At what time will the particle be traveling at 58◦ with respect to the horizontal?
  11. Oct 9, 2009 #10
    Sorry, I was not aware that 't' is time.
  12. Oct 9, 2009 #11
    Oh, it is very possible, ha ha. I am not good at this stuff at all.
  13. Oct 10, 2009 #12


    Staff: Mentor

    At t = 0, is the thing moving up or down? If it's moving up, vi will have a sign opposite to that of gravitational acceleration. Your equation in the first post suggests that the thing is moving up.
  14. Oct 10, 2009 #13
    I assume you are using equations for constant acceleration. You are going to need to break your equations into components: one for your vertical velocity and acceleration components and one for your horizontal components.



    You end up with two equations and two unknowns (V and t). You should be able to solve.
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