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1D motion Q

  1. Feb 22, 2005 #1
    Can someone just tell me where to start with this Q? I know it will have to solve some simult. eqs. to work it out.

    "A projectile is fired from the top of a cliff 150m high, and takes 10s to reach the base of the cliff after being fired."

    Find the initial speed.

    Thanks.
     
    Last edited: Feb 22, 2005
  2. jcsd
  3. Feb 22, 2005 #2

    Galileo

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    The only velocity component thsat is of interest is the vertical one.
    I think you have to assume the projectile is fired vertically, because else you cannot determine the initial speed since the horizontal component does not affect the falling time.
    Set up the equation of motion for the vertical component.
     
  4. Feb 22, 2005 #3
    Find the kinematics equation relating distance, initial velocity, time, and acceleration. Be careful with the signs of the quantities.
     
  5. Feb 22, 2005 #4

    dextercioby

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    Indeed the problem does not give the "range" (on the Ox=horizontal axis) of the projectile,so the answer should be given as:

    [tex]speed(t=0)=|\vec{v}(0)|=\sqrt{v_{0,y}^{2}+(\frac{R}{10})^{2}} \ [ms^{-1}] [/tex]

    ,where R is the horizontal range (in meters) and [itex] v_{0,y} [/itex] needs to be computed by the OP.

    Daniel.
     
    Last edited: Feb 22, 2005
  6. Feb 22, 2005 #5
    What are you saying :surprised . Since the range is not given, then we assume the motion is vertical. Moreover is given that the problem is 1D.

    Let's take the positive y-direction the one pointing downwards. So, g = +9.8m/s, s = +150 m, t = 10s.
    [tex]
    s = v_{0}t + \frac{gt^2}{2}
    [/tex]

    [tex]
    v_{0} = \frac{2s - gt^2}{2t} = -34 m/s
    [/tex]


    The negative direction indicates that the initial velocity is upward.
     
    Last edited: Feb 22, 2005
  7. Feb 22, 2005 #6

    dextercioby

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    Relax,where did you see such nonsense "A projectile is fired from the top of a cliff 150m high, and takes 10s to reach the base of the cliff after being fired." I find ridiculous that the movement is 1D,the projectile would come back into your head and,provided it does not explode,it would crush it...

    :tongue:

    Don't read more into it,than it's actually written...

    Daniel.
     
  8. Feb 22, 2005 #7
    Don't you ever use approximations, Daniel? :-p
     
  9. Feb 22, 2005 #8

    dextercioby

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    Sure,but it's not the case here.The text of the problem,even incomplete,is very clear and leaves no room for approximations/personal interpretations.

    Daniel.
     
  10. Feb 22, 2005 #9
    Thanks ramollari. It's that simple!

    The prob was that i was using distance, not displacement (wasnt considering them all as vectors).

    Btw, it is vertical motion -- i explicitly stated that it was 1d so there's no horizontal movement (even if there was, it wouldnt affect the initial speed). It's like the person leaned over and fired the bullet.

    Cheers
     
    Last edited: Feb 22, 2005
  11. Feb 23, 2005 #10
    That's a pretty slow bullet.
     
  12. Feb 23, 2005 #11
    wow i dont know what yall are doing, but i used this equation

    DeltaX = .5(A)(T)^2 + Vi(T)

    plug in the numbers....

    150 = .5(9.8)(10)^2 + Vi(10)

    dont read further if you dont want the answers....

    simplify.....

    150 = 490 + Vi(10)

    do some basic algebra....

    step1
    -340 = Vi(10)
    step2
    -34=Vi

    and its in the negative direction since its going down

    am i just crazy or something, but i think that equation works, i have no idea what you guys were doing, square rooting your stuff like that...unless you were using the:

    Vf^2 = Vi^2 + 2(A)(DeltaX) equation
     
  13. Feb 24, 2005 #12

    dextercioby

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    Incorrect,it's negative,since it's going UP,you've chosen the "g" with a positive sign,which automatically assumes the positive axis of coordinate to be pointing DOWN...:wink:

    Daniel.
     
  14. Feb 24, 2005 #13
    i was under the impression he was firing it downward...which would make it a negative direction if my starting point is the cliff top and distance above him is positive
     
  15. Feb 25, 2005 #14

    dextercioby

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    Then if you've chosen the "g" to be positive,your answer should be +34m/s.

    Daniel.
     
  16. Feb 25, 2005 #15
    i agree, it s 34m/s...
     
  17. Feb 25, 2005 #16
    either way the number is right, its just positive or negative by your opinion of where the startpoint is
     
  18. Feb 26, 2005 #17
    Opposite directions make a great difference! We have to know the direction of vector quantities, otherwise we go wrong in our calculations.
     
  19. Feb 26, 2005 #18
    gravity is just about always negative, it was just a typo that i left out the negative sign in the initial equation, so please just get over it and stop busting on me for it
     
  20. Feb 26, 2005 #19

    dextercioby

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    Gravity is a vector...Period.The sign (+/-) of the scalar depends on how u choose the axis perpendicular to the ground.Namely on how you chose its positive sense...

    Gravity (i.e.gravitational intensity) is therefore a vector.It's neither negative,nor positive.

    Daniel.
     
  21. Feb 26, 2005 #20
    Perhaps the cliff moved out of the way, Dexter :-)
     
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