A stone is released from an elevator going up with acceleration of $g/2$. What is the acceleration of the stone just after the release?

The answer is $g$. Shouldn't the stone carry the acceleration of the elevator and be $-g/2$?

  • $\begingroup$ Velocity is conserved, acceleration isn't $\endgroup$
    – seldon
    Oct 4, 2014 at 17:07
  • $\begingroup$ Can you please tell me, how velocity is conserved while acceleration is not. $\endgroup$
    – pcforgeek
    Oct 5, 2014 at 7:05

3 Answers 3


While the stone is still travelling on the elevator, there are two forces acting on it, the force from the elevator to the stone, as well as the weight due to gravity.

enter image description here

The moment the stone leaves the elevator, it becomes a free falling object. The elevator stops giving a force to the stone, and the only force remaining is its weight due to gravity.

enter image description here

From this you can see that as the only force is W = mg, the acceleration felt by the stone will be g.

While it is true it will be travelling upwards initially due to its momentum, its initial speed does not matter, as the only force that is acting on it would be force due to gravity, so its acceleration experienced will simply be $g$.

  • $\begingroup$ Perfect answer, how did you draw the graph? Online or by software? $\endgroup$ Oct 4, 2014 at 21:31
  • $\begingroup$ Thanks, I used Paint to draw a rough diagram to illustrate. $\endgroup$
    – t.c
    Oct 5, 2014 at 5:44

Acceleration only happens if a force is acting on an object. When you are holding the stone, the force which is accelerating the elevator is transmitted to the stone through your grip and you, the elevator and the stone form a single rigid collection of objects all accelerating at g/2. When you let go the only force left acting on the stone is gravity and it accelerates it at g.


While the speed and the position of an object change gradually (because of $x(t) = \int_0^t v(t') dt'$ and $v(t) = \int_0^t a(t') dt'$) the acceleration can change instantly therefore it doesn't matter for the acting force wether you just dropped the stone or not.

So while you're holding the stone it feels exactly the same acceleration as you do (the force from the gravity is canceled out by holding the stone)

$a = -\frac{g}{2}$

and as soon as drop the stone it's

$a = g$

If you drop the stone at $t = t_0$ a(t) would be:

$a(t) = -\Theta(t_0 - t)\frac{g}{2} + \Theta(t- t_0)g$

PS: $\Theta(x)$ is 0 if x is smaller than 0 and if it's equal or bigger than 0.


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