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Kinematics velocity Question

  1. Aug 11, 2015 #1
    1. The problem statement, all variables and given/known data
    The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 14 m/s. (a) What is the magnitude of the velocity of the projectile 1.2 s before it achieves its maximum height? (b) What is the magnitude of the velocity of the projectile 1.2 s after it achieves its maximum height? If we take x = 0 and y = 0 to be at the point of maximum height and positive x to be in the direction of the velocity there, what are the (c) x coordinate and (d) y coordinate of the projectile 1.2 s before it reaches its maximum height and the (e) x coordinate and (f) y coordinate

    2. Relevant equations
    Maybe will use component method


    3. The attempt at a solution
    JEdcQA0.png

    Have you tried anything yet?
    - First time encountering this kind of problem, but by using my knowledge in vectors(which is not that big, so don't expect.) maybe I can say I'll use 14 as my y component and use some data (which I do not know) to get the x component. Then I can use it to get the hypotenuse or the magnitude.

    It seems you only known a few about this problem?
    - Well sorry if I lack information about this problem, well this is why I posted in this forum right? For someone to lend me a hand.

    English is not my first language
     
  2. jcsd
  3. Aug 11, 2015 #2

    Titan97

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    Does the x-component of velocity of projectile change during its motions?
     
  4. Aug 11, 2015 #3
    Well I think so. Yes. If I'm wrong, please tell me why, I would love to know. Thank you!
     
  5. Aug 11, 2015 #4

    Titan97

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    What is the direction of acceleration of a projectile?
     
  6. Aug 11, 2015 #5

    Titan97

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    Hint: What you said in post 3 is wrong
     
  7. Aug 11, 2015 #6
    Oh okay? I think it is up to down only so maybe the x-component will not change?

    Then my drawing is wrong if that;s the case
     
  8. Aug 11, 2015 #7
    No , acceleration is in downward direction , but your diagram is still correct .
     
  9. Aug 11, 2015 #8

    Titan97

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    Leave the diagram for now. Good. The acceleration is indeed along y-axis only. Now, write the kinematic equations for displacement of projectile along x and y-axis separately.
    Take ux and uy as the initial velocity along x and y axis respectively and ax and ay as the acceleration. (You know the value of ax and ay)
     
  10. Aug 11, 2015 #9
    I think you mean this? q8GGHtZ.png
    the 2nd equation I assume?

    or you mean this
    Vx = VoCosΘ
    Vy = VoSinΘ

    ??
     
  11. Aug 11, 2015 #10

    Titan97

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    Take the third kinematic equation (given in blue).
    s=v0+½at2. Just write the equation along x and y axis separately.
     
  12. Aug 11, 2015 #11
    Sooo okay
    S=(14)+1/2(9.8)(1.2)^2
    Am I right?
    S=(14)+1/2(9.8)(-1.2)^2
     
  13. Aug 11, 2015 #12
    You haven't done any problem of this sort before ? Have you not even read about it before ?

    Projectile motion has different formulas , different conditions and you should read about it before trying any of it's questions . It isn't just a different kind of kinematics problem , it ( projectile motion ) consists of a whole set of different problems .
     
    Last edited: Aug 11, 2015
  14. Aug 11, 2015 #13
    But I'm sorta getting the idea of freefall. I will change a into g right? 9.8 m/s^2
     
  15. Aug 11, 2015 #14

    Titan97

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    Yes. I advice you to read about projectile motion before attempting this. Read this: https://en.wikipedia.org/wiki/Projectile_motion
     
  16. Aug 11, 2015 #15
    I sort of understand free fall which lead to projectile motion right?
    Look. I'm not totally clueless about the question. But a help is all I need, I'm not asking for answers.
    I'm a type of person who understands lesson via examples.

    This is my first time encountering this type of Q
    but this is not my first time answering a free fall question.
     
  17. Aug 11, 2015 #16

    Titan97

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    Please read about projectile motion and you will find that you could have solved the question yourself.If you are still not able to solve, then post your doubt on PF. I am helping you. I don't want to spoil the fun in solving this question.
     
  18. Aug 11, 2015 #17
    Okay done reading. So can you guide me in the question?
     
  19. Aug 11, 2015 #18

    Titan97

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    In the wikipedia page, the equation for displacement for projectile motion along x and y is given. Use that to find the time of flight. For that, substitute the value of y as zero since at the end of flight, the net displacement along y is zero.
     
  20. Aug 11, 2015 #19
    Please note typo. It should be s=v0t+1/2at2.

    With (x,y)=(0,0) at the top of the curve, apply this equation to the x and y directions separately.
     
  21. Aug 11, 2015 #20
    I got a and b last night. My answer for the magnitude is 18.2838

    I got it from sqrt(14^2+11.76^2)

    I got 11.76 from Vf = Vo-gt<- (9.8)(1.2)

    soooo next is c,d,e,f.. any help?
     
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