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Help With Projectiles Problem

  1. Apr 14, 2008 #1
    [SOLVED] Help With Projectiles Problem

    1. The problem statement, all variables and given/known data
    A car rolls down a slope at 30° to the horizontal, and reaches 40 m/s just before plunging off a cliff 180m high. What is its velocity just before hitting the bottom?


    2. Relevant equations
    [tex]U_H = vcos\theta[/tex]

    [tex]U_V = vsin\theta[/tex]


    3. The attempt at a solution
    I began the problem by finding intial velocity components:

    [tex]U_H = 40cos30^o[/tex]
    [tex]U_H = 34.6 m/s [/tex]
    [tex]U_V = 40sin30^o[/tex]
    [tex]U_V = 20 m/s [/tex]

    I then found the time the projectile takes to fall:

    [tex]s=ut+\frac{1}{2}at^2[/tex]
    [tex]-180=-20t-4.9t^2[/tex]
    [tex]4.9t^2+20t-180=0[/tex]
    [tex]t = 4.35s \mbox{(Using Quadratic Formula)}[/tex]

    The horizontal component stays constant at 34.6 m/s

    I then calculated the final vertical component

    [tex] V_v = U_v + at [/tex]
    [tex] V_v = -20 + (-9.8\times4.35) [/tex]
    [tex] V_v = 62.6 m/s \mbox {down} [/tex]

    I then constructed a vector diagram to calculate the resultant velocity. My working for the resultant velocity was as follows:

    [tex] V_R = \sqrt{62.6^2 + 34.6^2} [/tex]
    [tex] V_R = 71.5 m/s [/tex]

    [tex] tan \theta = \frac{62.6}{34.6} [/tex]
    [tex] \theta = 61.1^o [/tex]

    Hence, according to my working, the velocity just before hitting the bottom is 71.5 m/s at 61.1° below the horizontal. However, the answer given on the answer sheet to these questions is 65.7 m/s at 72.3° below the horizontal. I am very stuck because I am not sure how to get this answer. Can anyone see where I have gone wrong? Any help would be much appreciated. Thanks very much in advance.
     
  2. jcsd
  3. Apr 14, 2008 #2

    Hootenanny

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    You calculated the time incorrectly using the quadratic formula. However, there is no need to calculate the time of flight, instead, simply use a kinematic equation without time as a variable.
     
  4. Apr 14, 2008 #3
    Ok, I’m still learning this stuff too and apparently I don’t know it as well as I thought. Given this problem I would have solved it the same way, and came up with the same answer as technology.

    The only equation I can find that that relates velocity and time (it’s not in the book that I have) is

    [tex]

    v_y^2 = v_{yo}^2 - 2gy

    [/tex]

    [tex]

    v_y = \sqrt{(-20 m/s)^2 - 2(9.8 m/s^2)(-180)} \approx 62.7~m/s

    [/tex]

    [tex]

    t = \frac{v_{yo} - v_y}{g} = \frac{(-20 m/s) - (62.7 m/s)}{9.8 m/s^2} \approx 4.35 sec

    [/tex]
    the same as the quadratic formula. =(
     
  5. Apr 14, 2008 #4

    Nabeshin

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    I can see nothing wrong with what either of you did because I am repeatedly getting the same result. Used work-energy and kinematics, both obtained Vf to be 71.6. The answer key is probably incorrect.
     
  6. Apr 14, 2008 #5

    Hootenanny

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    Actually, scratch my previous post. You have calculated the time correctly, it was me that made the mistake (I has a sign wrong :blushing:). Your final answer and solution are correct.
     
    Last edited: Apr 14, 2008
  7. Apr 14, 2008 #6
    Actually my solution matches the answer key technology,
    heres how:

    Vy,f^2 = Vi^2(sin0)-2g(change in y)

    Vy,f^2 = 40^2(sin30)- 2(9.83)(180)

    Vy,f^2 = 1600(.5)+3538.8

    Vy,f^2 = 4338.8

    Vy,f = 65.7

    Don't second guess yourself hootenanny! :rolleyes:
     
  8. Apr 14, 2008 #7

    Hootenanny

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    Your solution is incomplete, what about the horizontal component of the velocity?
     
  9. Apr 14, 2008 #8
    you should have

    Vy,f^2 = 40^2(sin30)^2 - 2(9.83)(180)

    that will get you the 62.7 m/s for Vy,f that RyanSchw and Technology also got.
     
  10. Apr 14, 2008 #9
    Oh crap i see i left off the square on the (sin0)^2. Sorry hootenanny :redface:
     
  11. Apr 15, 2008 #10
    Thanks very much for the help guys, I think the answer key must be wrong. I appreciate the help.
     
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