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Easy kinematic problem I've suffered an hour for

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Metal ball is launch at 30 degrees from initial height of 1m with initial speed as 7.66m/s, find horizontal distance traveled

    2. Relevant equations



    3. The attempt at a solution

    Honestly I have no idea.
     
  2. jcsd
  3. Sep 16, 2009 #2

    rl.bhat

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    Homework Helper

    Go through any text book or hyper Physics site and collect the information about the projectile motion. If you have any doubts, please feel free to ask us. Before that you have to show your attempt.
     
  4. Sep 16, 2009 #3
    you have to use the kinematics equation s= u*t - 0.5gt^2 in vertical and horizontal directions. u is the initial velocity and s is the distance travelled. g is the gravi . acc = 9.8 ms-2 , when it reaches floor level it has travelled -1m from the place of launch. use above equation for vertical direction with s= -1m , and u=7.66* sin30 , you have to resolve for the vertical component of launch speed, now when you plug in this data in above equation you will get a quadratic equation in t you can find two roots to this equation , one root will be (-) so you can take the positive root, this is the time it takes to get to the floor level, now think of the horizontal displacement there is no acceleration in that directin so the resolved horizontal speed ( 7.66*cos 30) will remain unchanged so the horizontal dispacement = horizontal speed * time it takes to getto the floor !!!!!
    good luck
     
  5. Sep 16, 2009 #4


    check my previous posts, I am not a leech
     
  6. Sep 17, 2009 #5
    What's your previous attempts though?

    Did you try breaking it down into the x- and y-components?

    Even a sit down with a sketch to familiarise yourself with the problem helps.

    :smile:
     
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