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Need Help with this Question

  1. Oct 3, 2005 #1
    Need Help with this Question!!!!!

    i know it sounds simple but everytime i find myself lost.

    a projectile is launched with an initial speed of 22m/s 47deg above the horizontal from a 38m cliff, determine the displacementof the object 2.1s into its trajectory.

    do i find the displacement using dx=vit+0.5t2?

    please help!
     
  2. jcsd
  3. Oct 3, 2005 #2

    kreil

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    Gold Member

    all you need to do is use trig to separate the initial speed they give you into x and y components, and then use d=vt first using the horizontal velocity and then the vertical velocity to find how far it moves in each direction, then you construct a triangle with these 2 sides and solve for the 3rd side, which will be the displacement of the object.
     
  4. Oct 3, 2005 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    "do i find the displacement using dx=vit+0.5t2?"

    I can't answer that because I am not sure what you are saying. It certainly is true that dy= (viy)t- (0.5)gt2+ y0 where dy is the change in vertical height, viy is the vertical (y) component of initial velocity, g is the acceleration due to gravity,y0 is the initial height, and t is the time in seconds after launching.

    Do you see the difference? I'm not complaining about your not using html tags (though even t^2 would be easier to read than "t2") but about the fact that you did not explain what your symbols meant. Certainly, I would have been inclined to think of dx as horizontal distance rather than vertical, in which case there is no acceleration.

    Since the problem asks for "displacement", the vector change from the initial position, the initial height, y0= 38 m is irrelevant.
    Use dx= vixt and dy= dy= (viy)t- (0.5)gt2+ y0 to find the displacement.
     
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