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Time for liquid ejecting from a hole in tank to reach ground

  1. Dec 3, 2011 #1
    Hi all,

    I was trying to find the time required for liquid ejecting from a hole in a tank to reach ground level. The equation given in one of the books is as follows:

    t = √(2h/g) or sqrt of (2h/g).

    where h stands for the height of hole in tank shell from the ground level.

    This equation is true when the body is free falling with zero initial velocity. However, my question is whether the equation is valid when we are considering a liquid ejecting from the hole (which will definitely have initial velocity due to head of remaining liquid in the tank)?

    Thanks in advance.
  2. jcsd
  3. Dec 3, 2011 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Will a horizontal initial velocity matter?
  4. Dec 3, 2011 #3
    Hi Doc Al,

    Thanks for the reply. You have infact echoed my concern too. I think it matters. But when the text book does not account for it, I am confused.

    To give a clearer picture, the objective of the problem is to determine horizontal distance 'S' by which the liquid will be thrown from the hole.

    In the textbook, once the time 't' is found out, they use this value of 't' in the following equation to determine 'S':

    S=v*t where v is the velocity of the liquid at the hole (this velocity is determined using the energy conservation equation)

    I was trying to analyse the same condition using another example. If in Case 1 a bullet is fired from a gun and in Case 2 a similar bullet is dropped from the same height of the gun, both at the same time, will both bullets reach ground at same time? I dont think so...but I cant prove it.

  5. Dec 3, 2011 #4

    Doc Al

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    Staff: Mentor

    Realize that you can treat vertical and horizontal motions completely independently. What's the equation for vertical position as a function of time for a projectile? Does the horizontal motion matter?
  6. Dec 3, 2011 #5
    Hi Doc Al,

    thanks for the thoughts. I visited the following website: http://electron9.phys.utk.edu/phys135d/modules/m3/Projectile%20motion.htm [Broken]

    It says that we can treat motion in two dimensions.

    Vx=V0x and Vy=Vy0-gt


    x = Vx0 t and y = Vy0t - 1/2 g t^2


    t = sqrt(2y/g)

    With this case, it can be concluded that both the bullets must reach ground at same time.

    Am I correct?.

    Thank you very much
    Last edited by a moderator: May 5, 2017
  7. Dec 3, 2011 #6

    Doc Al

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    Staff: Mentor

    Yes, you are correct. The time it takes for an object to reach the ground depends only on its initial vertical speed (and distance from the ground, of course). Horizontal speed doesn't matter. The two bullets will hit the ground at the same time. (Ignoring complications of air resistance, of course.)
  8. Dec 3, 2011 #7
    Great. Thanks :)
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