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Prove that an object, if thrown up under gravity

  1. Jun 29, 2009 #1
    Can you show me how to prove that an object, if thrown up under gravitational force, will take the same time to reach the apex of it's trajectory as it takes to fall back to it's original position? (ignore air resistance)
  2. jcsd
  3. Jun 29, 2009 #2


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    You are then assuming constant acceleration. You should be able to write down the standard formula for the height of the object at any time t, for any initial speed v0. Find the time until it reaches its highest point and the time until it comes back to the original height. Observe that the first time is exactly half the second time.
  4. Jun 29, 2009 #3
    well its easy
    For parabolic path
    because the time to rise and fall is affected by gravity (and gravity affects upon vertical velocity)
    Break the velocity u into ucosx and usinx (x is angle)
    here is vertical velocity is usinx
    since it is affected by gravity its value decreses in maximum height that is V=o
    and a = -g
    that is thime to rise T1= o-usinx/ -g = usinx/g
    and similarly after reaching the highest point than the gravity favours downward velocity and now velocity goes to increase and becomes maximum at the hitting the ground
    At u=o and v= usinx
    T2 = usinx-o/g
    =usinx/g T1=T2
    hence proved
  5. Jun 30, 2009 #4
    If we consider forces F(x) that depend only on position x, then Newton's law takes the form:

    x'' = F(x)

    This equation is invariant under time reversal t -> -t since by the chain rule on the LHS we have minus time minus is plus, and the RHS is manifestly invariant. Therefore under these conditions the laws of mechanics are invariant under time reversal, from which the proposition stated by the OP follows as a corollary.
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