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Basic Kinetics Question [Secondary]

  1. Sep 7, 2006 #1
    Okay I suck at physics and if someone could help me out it would be greatly appreciated.

    • A ball is projected horizontally from the edge of a table which is 0.8 m high, as shown below. The ball lands 0.2m from the table
    • How long does the ball take to fall to the ground?

    Worked it out to be 0.4s using s = ut + 1/2 at^2

    0.8m = 1/2 * 9.8t^2
    1.6/9.8 = t^2

    • Hence show that the ball was projected with a horizontal speed of 0.5ms

    but whenever I work this out I get 4m/s. :(

    v = u + at
    v = 9.8 * .4
    v = 4m/s

    Dont think this place is really for Secondary level Physics but I couldnt find anywhere else :cry:
  2. jcsd
  3. Sep 7, 2006 #2

    Doc Al

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    Staff: Mentor

    The horizontal motion is not accelerated!
  4. Sep 7, 2006 #3
    ok... the problem is you are using the vertical acceloration as the horizontal acceloration. In the horizontal direction the ball is NOT accelerating. you know that it took 0.4 s to land and in that time it went a horizontal distance of 0.2 m.... speed = distance/time
    Last edited: Sep 7, 2006
  5. Sep 7, 2006 #4
    Think of the different forces affecting the object in the x and y direction, each by itself.
  6. Sep 7, 2006 #5
    I don't get what you mean? :-/

    Could you direct me to some notes or something...I've tried the ones from hsn.uk.net but I cant figure it out from them

    Edit : thank you dmoravec, I tried that formula as well the first time but I was putting in the wrong numbers :-p
    Last edited: Sep 7, 2006
  7. Sep 7, 2006 #6
    think of it this way... once the ball leaves the table what is happening? gravity is pulling it straight down, but there is nothing pulling it in the horizontal direction. Therefore with no force in the horizontal direction there is no acceleration.
    Last edited: Sep 7, 2006
  8. Sep 7, 2006 #7


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    Homework Helper

    (g*t^2)/2 = 0.8 => t = 0.4 s ;
    v*t = 0.2 => v = 0.2/0.4 = 0.5 m/s
  9. Sep 7, 2006 #8
    Yeah thanks all, was using the Horizontal component when I should of been using the vertical :-/

    2 more questions please :smile:

    1) Can someone point me to a site that explains Horizontal and Vertical speed/time graphs. I've tried google but found nothing

    2) How do I find the horizontal velocity of projection/vertical velocity it has on reaching the ground.
    Would the horizontal velocity be found by distance over time?
    i.e 25m/2.5 = 10m/s

    Would the vertical velocity be t * -9.8m/s?
    ie. 2.5seconds x -9.8m/s = -24.5m/s

    cheers :!!)
    Last edited: Sep 7, 2006
  10. Sep 7, 2006 #9
    Bump, edited the last post a bit so that I might get some answers :-(
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