My book states that the distance traveled by a particle thrown upward in the last second is independent of it's initial velocity because the the object travels the same distance as a freely falling body in $1^{st}$ second. But I have two concerns related to this:

  • Can't it be extended to the case for the distance traveled in $n^{th}$ second? My logic is that since the motion of the particle would be the same in forward as well as backward direction of time therefore distance traveled by it in $n^{th}$ second must be independent of the initial velocity.

  • Suppose that for the course of the motion the forces acting on it ceases to exist, then the distance traveled by the particle in $n^{th}$ second must be "$u$" meters. ( $u$ being the initial velocity). Hence,I conclude that even the case for accelerated motion must depend on initial velocity.

I am currently confused between the two (the fact that it should deped and it should not). So my question is would it depend on initial velocity or not? (If so how? Why?)

  • $\begingroup$ Are you concerned of aerodynamic effects? $\endgroup$ Nov 11, 2019 at 18:05
  • 1
    $\begingroup$ Note that the last second may not correspond to any nth second, because the nth seconds are counted from where t is zero, but the last second is measured from where v is zero (which may be at some fraction of an nth second.) $\endgroup$
    – R.W. Bird
    Nov 11, 2019 at 20:01
  • $\begingroup$ For an object thrown up (without friction), the motion coming down is symmetrical with the motion up. So the motion in the first second coming down from the top, (where the vertical velocity is zero) does match the motion in the last second going up (approaching a velocity of zero). $\endgroup$
    – R.W. Bird
    Nov 11, 2019 at 20:15

1 Answer 1


I am assuming that the body is in free fall and no aerodynamic effects are involved in the system.

Let the body be thrown upward with a velocity $u$,then it would take $\frac{u}{g}$ seconds to reach the maximum height,then distance traveled in the last second will be equal to the displacement in the last second which is equal to:- $$y_{\frac{u}{g}}-y_{\frac{u}{g}-1}$$ Using the equation of displacement for a body thrown upward, $$y_{t}=ut-\frac{gt^2}{2}$$ I think that the notations would be known to you if there is a confusion then ask me in the comment section.After simplification you will get:- $$y_{\frac{u}{g}}-y_{\frac{u}{g}-1}=u+\frac{g-2u}{2}$$ $$y_{\frac{u}{g}}-y_{\frac{u}{g}-1}=\frac{g}{2}$$ Now it is clear from the calculations that the distance traveled in last second is independent of initial velocity.


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