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Problem with finding time from projectile motion problem

  1. Jul 26, 2012 #1
    1. The problem statement, all variables and given/known data
    A projectile is fired from a cliff 125m off the ground at 37° from the horizontal at 65 m/s. How long will it take to hit the ground?


    2. Relevant equations
    y= yo+(Vyo)(t) - 1/2(g)(t^2)
    quadratic formula


    3. The attempt at a solution
    substituting in equation....

    (-125m)= 0 + (65m/s)(t) - 1/2 (9.8 m/s^2)(t^2)

    (4.9 m/s^2)(t^2)- 65 m/s (t) -125m =0

    rewrite in quadratic to solve for t

    t=((65 m/s) +/- √((65)^2-4(4.9 m/s^2)(125)))/ 2(4.9m.s^2)

    t = 10.9 s

    answer in back of book is 10.4 s

    not sure if using proper equation or totally missing something.
     
  2. jcsd
  3. Jul 26, 2012 #2

    tiny-tim

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    welcome to pf!

    hi coneheadceo! welcome to pf! :smile:

    erm :redface:

    where's the 37° ? :wink:
     
  4. Jul 26, 2012 #3
    Heyas Tim,

    The 37° is the angle the projectile is fired at from the top of the cliff with respect to the horizontal.
     
  5. Jul 26, 2012 #4

    tiny-tim

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    yes i know! :smile:

    but it's not in your equations!! :wink:
     
  6. Jul 26, 2012 #5
    So would I go for something like this...

    To find the Vyo = (65 m/s)Sin 37 and Vxo = (65 m/s)Cos 37 instead of just using 65 m/s?

    so then...
    (-125m)= 0 + [(65m/s)sin 37](t) - 1/2 (9.8 m/s^2)(t^2) for the y component of time

    and
    (-125m)= 0 + [(65m/s)cos 37](t) - 1/2 (9.8 m/s^2)(t^2) for the x component of time?

    Then do I just add them after using the quadratic for each?
     
  7. Jul 26, 2012 #6

    tiny-tim

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    yes :smile:
    yes :smile:
    erm :redface:

    what is the acceleration in the x direction? :wink:
    hmm … draw a diagram, and then think

    what do the x and y equations each tell you?​
     
  8. Jul 26, 2012 #7
    So should my acceleration be (+) and I would end up with an equation....
    (-125m)= 0 + [(65m/s)cos 37](t) +1/2 (9.8 m/s^2)(t^2)

    I drew a diagram and am at a loss for what each equation "should" be telling me.
     
  9. Jul 26, 2012 #8
    As tiny-tim already said, what's the acceleration in x direction? g acts vertically downwards, is their any component of g in the x direction?

    Show us the diagram you have drawn.
     
  10. Jul 26, 2012 #9
    This is wrong. The time it takes for the projectile to land depends only in the y direction. Therefore, your equation should reflect only the motion of the projectile in the y direction. your velocity should therefore be 65sin(37°)m/s instead of simply 65m/s as you have here. This is the mistake which accounts for the error in your answer.
    I worked it out using 65sin(37°)m/s as the velocity and got 10.38s for the time.
     
  11. Jul 26, 2012 #10
    I'll have to recheck my math. Thank you for the correction swill777!
     
  12. Jul 27, 2012 #11
    Nope.... banging my head against the wall here... I'm missing something with my math.
     
  13. Jul 27, 2012 #12
    HALLELUJAH!!!!


    Nevermind now people. Go back to drinking coffee as I needed a whole ton to realize.....


    my calculator was set for answers in RAD instead of DEG

    http://roflrazzi.files.wordpress.com/2010/04/129164952429291808.jpg [Broken]

    Thank you for your patience and help!
     
    Last edited by a moderator: May 6, 2017
  14. Jul 27, 2012 #13

    PeterO

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    It is always a good idea to enter sin(30) into your calculator before each set of calculations just to check it gives you 0.5 - indicating if it is actually set to degrees.
     
  15. Jul 27, 2012 #14

    Yup, the degree/radian mix up will get ya. I've had to make it a habit to check my setting every time I deal with trig. Glad you got the right answer, though!
     
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