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Acceleration from a position f(t), and a projectile motion problem (height, time)

  1. Oct 2, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    Hey guys, hello. There is a system belongs in internet called LON-CAPA, and my teacher told that homeworks are done via that system. So I logged in, the questions are not that difficult to solve. But system doesn't accept some of my answers which i'm so sure that being true. So I'm gonna write 3 easy questions for you, I'm just wondering if I did a mistake or not. Here we go.

    1 - The position of an object as a function of time is given as x = At3 + Bt2 + Ct + D. The constants are A = 2.31 m/s3, B = 1.07 m/s2, C = -4.91 m/s, and D = 3.45 m.
    Q-1 What is the acceleration of the object at t = 0.357 s?
    (MY ANSWER IS 7,088 m/s^2)

    2- A ball is thrown straight upward in the air at a speed of 17.1 m/s. Ignore air resistance.
    Q-2.1 What is the maximum height the ball will reach?
    (MY ANSWER IS 14,92m)

    Q-2.2 How long will it take to reach 2.31 m above its initial position on the way down?
    (MY ANSWER IS 3,34s)

    Thank you guys for you answers already!
     
  2. jcsd
  3. Oct 2, 2016 #2

    haruspex

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    I agree with your first two answers but not the third. Please post your working.
     
  4. Oct 2, 2016 #3
    I solved this in two steps. First, I calculated the landing time.

    I calculated g=9.8 m/s^2.
    V= Vo+g*t
    0=17.1+(9.8)*t
    tlanding=1.744 s (landing time)

    Then, i calculated the time between the maximum height position to 2,31 m above ground position.


    maximum height=14,92 m (Found it another question, which you said that you found it too)
    14.92-2.31=12.61 m to go down
    X=Vo*t+(1/2)*g*t^2
    12.61=0+(1/2)*9.8*t^2
    t=1.6 s

    t1+t2=3.34 s (whole process)
     
  5. Oct 2, 2016 #4

    haruspex

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    Ok, I found my mistake, but it is nearer to 3.35. You may have got a bit of rounding error because you calculated two stages and added them together. That is not necessary. You can solve it in one step with s=ut+1/2 at2. You just have to pick the larger root of the quadratic.
     
  6. Oct 2, 2016 #5
    So in conclusion, you tell me that you found around 3.34,-3.35. So system must accept my answers cause the answer don't have to be the absolute true. In my other calculations, it accepted answers like 7,1 which's true answer is 7,15.
     
  7. Oct 2, 2016 #6

    gneill

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    [Mod note: Please note that I have changed the thread title to better reflect the thread content. Pleas for help are not descriptive of the physics to be discussed and not allowed as thread titles]
     
  8. Oct 2, 2016 #7

    haruspex

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    Yes.
     
  9. Oct 3, 2016 #8
    Thanks, really appreciate it.
     
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