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Probably an easy kinematics question

  1. Dec 10, 2006 #1
    1. The problem statement, all variables and given/known data
    If a ball is fired upwards on an unknown planet at 17 m/s from a gun being held 32m above the ground, and it takes 20 seconds before the ball hits the ground, what is the magnitude of the planets 'g' value (acceleration)?


    2. Relevant equations

    kinematic eq'ns

    3. The attempt at a solution

    i really don't understand this problem, can someone give me a hint?? how would i find the final position for when the ball is fired up?
     
  2. jcsd
  3. Dec 10, 2006 #2

    cristo

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    what can you say about the velocity of the ball when it reaches its maximum height?
     
  4. Dec 10, 2006 #3
    the velocity at the maximum height would be 0m/s
     
  5. Dec 10, 2006 #4
    i don't know how to find x without being given an acceleration or time.. maybe i'm missing an eq'n?
     
  6. Dec 10, 2006 #5

    cristo

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    well, what equations do you have, and what have you tried? If you post your attempts, however small they may be, then someone will be able to point you in the correct direction.

    Remember, you only need to find the acceleration due to gravity, not the vertical height
     
  7. Dec 10, 2006 #6
    Consider one dimension only,

    the acceleration = -g

    velocity=-gt+initial velocity

    displacement=0.5*gt2-t*initial velocity

    those are the "kinematic eq'ns" you might be looking for. I don't think you really need to know velocity at the top, cristo was pulling your leg! provided you can rearrange and plug things in you;ll be fine.

    What is the displacement you are looking for?
     
    Last edited: Dec 10, 2006
  8. Dec 10, 2006 #7

    cristo

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    Yea, I realised that.. its a lot simpler to just consider the complete motion!
     
  9. Dec 10, 2006 #8
    arghhhhhh i still don't understand any of this :uhh: ... maybe thats why i'm a biology student haha :tongue2:

    anyways, i'm stuck on the idea of finding the maximum height ... and this would give me acceleration i dont know what the hell im doing haha
     
  10. Dec 10, 2006 #9
    I can't help you any more without actually doing it for you.

    To simplify matters don't consider what happens in those 20 seconds before the ball hits the ground. Just use the last kinematic equation I gave you, the displacement is relative to where the ball was launched (taking a "positive upwards" convention, the ball hits the ground at -32m). You've got one unknown and that's g, rearrange and solve....
     
  11. Dec 10, 2006 #10
    maybe im reading the question wrong.. but im assuming the 20seconds is the time the ball leaves the gun until it reaches the ground? or is the 20 seconds from the maximum height until it reaches the ground? anyway the equation you gave has 2 unknowns so therefore i dont see how i can solve it.. yes i suck at math
     
  12. Dec 10, 2006 #11
    your assumption about the 20 seconds being the time from when the ball leaves the gun is correct!

    you know the displacement.
    you know the time.
    you know the initial velocity.

    all you have to do is find g.
     
  13. Dec 10, 2006 #12
    how do i know displacement if i dont know the maximum height???
     
  14. Dec 10, 2006 #13

    cristo

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    You're shooting a ball from a vertical height of 32m directly up in the air. If you consider the whole motion of the ball, to when the ball hits the ground, what do you think the total displacement is? Billiards gives you a hint (well, the answer!!)

    To clarify this: consider the point from which the ball is released to be the origin. Let h represent the maximum height, measured from this origin. Then, the total displacement is h+ (-h) -32 (since the ball is released from 32m below the origin)
     
  15. Dec 10, 2006 #14
    You can solve this problem with a single kinematics equation. This equation has the following: initial height, initial velocity, and time. You are given all of those variables. This equation also includes (g) or gravity. What is this equation? :smile:
     
  16. Dec 10, 2006 #15
    well i used the equation x=x(initial)+v(initial)t+0.5at^2 and i got the answer of 1.24m/s^2 is that right?
     
  17. Dec 10, 2006 #16

    Yeah that is the correct equation however I think you should use the variable "y" instead of "x" because height is vertical just like the y-axis on a coordinate plane.

    I didn't get that answer, I think you subsitute some numbers in the wrong place. Please write out the equations with all the numbers in the correct places for me. I'll see what you did wrong from there.

    Can anyone confirm if his procedure is correct? I'm a physics student myself and this is my first year as a sophomore taking it. So I'm not that experienced as many other people on this forum.
     
  18. Dec 10, 2006 #17
    x=0.5g(t^2)-t(v1)
    -32=0.5g(20^2)-20(17)
    -0.5g(400)=32+(-20(17))
    -0.5g(400)=-308
    g(400)=616
    g=616/400
    g=1.54

    sorry i think my last one was wrong this one might be too but i got 1.54 this time
     
  19. Dec 10, 2006 #18

    cristo

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    The problem is 1d, so x is fine as a choice for a variable

    What does x correspond to in this equation. If it corresponds to x(final) then yes, the equation is correct, supposing that x(final) = 0, and x(initial) =32. (or x(final)=-32, x(initial)=0)
     
  20. Dec 10, 2006 #19

    cristo

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    a minus sign has crept in on the first line in front of (v1)t. change this to a plus, and the answer will be correct
     
  21. Dec 10, 2006 #20
    i assumed xinitial was 0m and xfinal was -32m... i just am having trouble visualizing how this is calculated without knowing how high the object actually reaches


    so g=-1.86 ?
     
    Last edited: Dec 10, 2006
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