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Homework Help: Quadratic solving for T

  1. Apr 15, 2014 #1
    15=35 sin Theta*t -4.9t^2

    I have tried solving for t, I do not know how to...

    please help

    thank you
  2. jcsd
  3. Apr 15, 2014 #2
    What is Θ? (If you click on the "Go Advanced" button you will be able to put in the Greek letters as needed it also allows you to use superscripts.

    You did not use parenthesis so I can tell you mean
    15=35 sin (Θ)*t -4.9t2, or
    15=35 sin (Θ*t) -4.9t2, or
    15=35 sin (Θ*t -4.9)t2, or
    15=35 sin (Θ*t -4.9t)2, or
    15=35 sin (Θ*t -4.9t2), or

    The answer to your question depends on that. Bottom line: You need to learn to use parenthesis as needed.
    Last edited: Apr 15, 2014
  4. Apr 15, 2014 #3
    15=35 sin (Θ)*t -4.9t^2

  5. Apr 15, 2014 #4
    What is the solution to the equation at2 + bt + c = 0 ?
  6. Apr 15, 2014 #5
    ((-b(+-)sqrt(b^2 -4ac))/2a
  7. Apr 15, 2014 #6
    I know the quadratic formula, but how will I do it if j have theta as well?
  8. Apr 15, 2014 #7
    b = 35 * Sin(Θ) in this case. It is a symbolic formula, if it is a quadratic, it is valid.
    Just make your equation "fit" the format of the formula and you will see what is a, what is b and what is c.
    Then solve it.
  9. Apr 15, 2014 #8

    x= [itex](4.9 \pm \sqrt{24.01+2100 sin\vartheta}/(70 sin \vartheta)[/itex]
  10. Apr 15, 2014 #9
    It doesn't look like you applied the formula correctly. What did you use for a, b, and c?

  11. Apr 15, 2014 #10
    15=35 sin (Θ)*t -4.9t2

    a=35 sinΘ
  12. Apr 15, 2014 #11
    This is wrong. You posted the correct formula above but identified wrong values for a, b and c.
    Also, you missed a parenthesis.
  13. Apr 15, 2014 #12
    a is the coefficient of the t^2 term, b is the coefficient of the t term and c is the independent term. Try again. Why don't you rearrange it as I suggested, it is a good practice if a, b and c don't seem ridiculously obvious.
  14. Apr 15, 2014 #13

    So I have it now, how would i simplify the following (I have subbed it into x=35cos[itex]\vartheta*t[/itex])?

    x=35cos[itex]\vartheta[/itex]([itex]-35 sin\vartheta \pm \sqrt{(35sin \vartheta)^2 -294}/-9.8[/itex]

    I tried to put in the parentheses but it would not let me
    Last edited: Apr 15, 2014
  15. Apr 15, 2014 #14
    You need to get the algebra right before we can help you simplify anything.

  16. Apr 15, 2014 #15
    all terms except 35cos theta, should be divided by -9.8. I tired to show the parentheses but it wouldn't work
  17. Apr 15, 2014 #16
    I may be missing something, but why is there a cos(theta) anyways?
  18. Apr 15, 2014 #17
    Im subbing it into another equation
  19. Apr 15, 2014 #18
    Be clearer, are you trying to do:
    1)[itex] cos{(\theta \cdot t)}[/itex]
    2)[itex] cos{(\theta)} \cdot t[/itex]

    Also, it seems to me that your solution for x is still not correct. Check your signs.
  20. Apr 15, 2014 #19
    The second option :)
  21. Apr 15, 2014 #20
    As he said, redo it. From what I can see, your signs are wrong.
  22. Apr 16, 2014 #21
    It seems you're solving a problem of projectile motion. Is this correct? Would you post the problem so we can help you more effectively?
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