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Help needed

  1. Feb 1, 2004 #1
    Can anyone help me solve the following question:
    A particle is moving vertically at 4.8 m/s and accelerating horizontally from rest at 1.7 m/s2. At what time will the particle be traveling at 32 degrees with respect to the horizontal? I have got various answers such as 0.894s, 0.44704s, 0.383369s, 0.54934s, 0.441899 and 0.88379s and none of them are correct according to the automatic grading website my professor use for our physics class. Appreciate!!!
     
  2. jcsd
  3. Feb 1, 2004 #2

    enigma

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    Hi Xamfy,

    welcome to the forums!

    Post your work so we can help find where the problem is
     
  4. Feb 1, 2004 #3
    I assumed that
    Vy = 4.8 - 9.8t and
    Vx = 1.7t
    tan(32) = 0.6249 = Vy/Vx, by solving the equation, I got t = 0.4419, which is not the correct answer. I also try to assume the particle move downward so that
    Vy = 4.8 + 9.8 t and
    Vx = 1.7t
    tan(32) = 0.6249 = Vy/Vx, which gave me a negative number.

    I just tried to assume that the particle is moving upward and the gravity is positive. This produced the following function--

    Vy = -4.8 + 9.8t
    Vx = 1.7t

    0.6249 = (-4.8 + 9.8t)/1.7t, Thus,
    t=0.59259 s

    This is a new value I haven't confirmed yet. I guess I was confused by the original moving direction of the particle. I am also not sure about the sentence "and accelerating horizontally from rest at 1.7 m/s2" If the particle is moving vetically at 4.8 m/s, how can it be at rest when horizontal acceleration is mentioned?


    Thanks for the reply.
     
    Last edited: Feb 1, 2004
  5. Feb 2, 2004 #4
    the formula required here is

    [tex]\tan32 = \frac{|V_y|}{|V_x|}[/tex]
     
  6. Feb 2, 2004 #5
    Thanks, but other question is what should be the Vy, 4.8 - 9.8t, 4.8 + 9.8t, or simply 4.8 itself.
     
  7. Feb 2, 2004 #6

    HallsofIvy

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    I would take "accelerating horizontally from rest at 1.7 m/s2" to mean that the particle was initially not moving horizontally. That's basically what you did.

    The particle is going at 32 degrees to the horizontal going upward when t= 0.44 seconds and going downward at 0.549 seconds which are answer you say have been rejected.

    Since the problem gives a horizontal acceleration are you sure that you are supposed to include gravitation? It's a bit of a stretch but since you have already "tried" several answers, see what the automatic grading does if you just drop the "9.8".

    By the way- some of those can be very sticky when it comes to accepting "just the right" answer. What were you told about number of decimal places to give?
     
  8. Feb 2, 2004 #7
    It Vy= 4.8-9.8t is correct if initially it is thrown upwards and including g. it would simply be 4.8 if neglecting g.

    Conceptwise the above are true whatever the ans is maybe it demands correct decimal places
     
  9. Feb 2, 2004 #8
    Thanks,, himanshu121 and HallsofIvy. I am running out of time for submitting the correct answer to this question. The posted answer from the professor included the neglect g, which means Vy is 4.8 only. This is a good lesson for me to know there are many possible sticky questions existing.
     
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