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Horizontal velocity experiment

  1. Jan 1, 2004 #1


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    Hai all,
    In projectiles the horizondal component of their velocity remains constant and veritical compont changes as it gets accelarated.So with this, I think we can find an experiment that gives the accelaration due to gravity(g) with out a clock.i.e if we are droping a ball from a table of height 5m with some horizondal velocity(its veritical component will be zero initialy)it reaches a horizondal distance of 10m at the bottom from the table.If we repeate this experiment with a table of height 10m and with same ball and same horizondal velocity the horizondal distance traveled is 13m.Then i don't know is there any way to calculate the accelaration due to gravity at that place.Any way it is clear that ball travels its second 5m (vertical) faster in its 10m drop.
    Anyway if a clock is a must for calculating 'g' please say why?

    **These values are not correct...

  2. jcsd
  3. Jan 2, 2004 #2


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    What you are saying is this: Have an object on a 5 m high table, and give it an initial velocity v0. Horizontal acceleration is 0 and vertical acceleration is -g. At time t, the horizontal speed is v0 and vertical speed is -gt. At time t, the horizontal distance from the table is v0t and height is 5- (g/2)t2. The object hits the ground when
    5- (g/2)t2= 0 or t= √(10/g) and will hit the ground at distance v0&radic(10/g). Let's assume that is 10 m as you say.
    If the height of the table is 10 m (Gosh, those are high tables!),
    the same calculation gives v0&radic(20/g) as the distance at which the object hits the floor. Assume that is 13 m.
    We now have v0&radic(10/g)= 10 and v]0&radic(20/g)= 13. From the first equation, 10/g= 100/v02 so g= v02/10. From the second equation, v02= 169g/20;. That gives g= (169g)/200.
    Hmmm, the "g"s cancel out! Ain't that a kick in the head!

    (I also suspect that your "13 m" and "10 m" values are not correct!)
  4. Jan 3, 2004 #3
    Theoretically, there is no need to compare it to a second experiment (although it is a good idea to make sure that there wasn't an error!).

    If the experiment is performed in a vaccuum so there is no air resistance:

    For the horizontal velocity, the following equation applies:
    d = vxt
    So, t = d/vx

    You know d and vx.

    Replace t with this value in the equation for vertical postions

    y = y0 - 1/2gt^2
    y = y0 - .5g(d/vx)^2

    Pluge in the values. Let c = (d/vx)^2.
    0 = 5 - .5gc
    10/c = g
  5. Jan 5, 2004 #4
    I may have not understood your question. Do you know the horizontal velocity, or do you just know that it is the same for both experiments?
  6. Jan 6, 2004 #5


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    Don't know velocity

    I tried it with out knowing horizontal velocity.If I know that it is very easy as u said.I asked for some kind of geometrical methods for calculation,by comparing two parabolas made by falling balls.But now i think it is not possible.
    Thank you.

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