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Time of ascent = Time of descent. My question is, In case of time of ascent gravitation pull is not favouring the motion (retardation occuring) but in case of time of descent gravitation pull is favouring motion (acceleration occuring). Then why time in both case is same? (π™‹π™§π™€π™«π™žπ™™π™šπ™™ π™©π™–π™ π™š π™–π™žπ™§ π™§π™šπ™¨π™žπ™¨π™©π™–π™£π™˜π™š π™£π™šπ™œπ™‘π™žπ™œπ™žπ™—π™‘π™š).

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  • $\begingroup$ You may be interested to look up Conservative Force Field, which gravity is in your example. In such a field, kinetic and potential energy are exchanged back and forth in a symmetrical manner, with no losses. $\endgroup$
    – RC_23
    Aug 6 at 16:36

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If you consider a homogenuous gravitational field (thus where the acceleration due to gravity does not depend on how high you are), then you simply have two motions which add up to the overall motion:

The linear motion due to the throw: $$s_{lin} = v\cdot t$$

Superimposed is the accelerated motion due to gravity: $$ s_{grav} = -\frac{g}{2}t^2$$

Now you can simply look at this from a perspective of symmetry:

You start with an initial velocity, and gravity's constant acceleration first has to stop this. And then the exact same path has to be taken backwards (and the stone will arrive with the same velocity on the ground (but moving downward) as it was thrown upward initially.

Mathematically we can look at this way: First we calculate the time until the stone returns to its starting place, thus the whole up and down motion, thus overall movement is 0:

$$s_{lin} + s_{grav} = 0\\v\cdot t - \frac{g}{2}t^2 = 0\\t^2 - \frac{2v}{g}t = 0\\ t_{1,2} = \frac{v}{g} \pm \sqrt{\left(\frac{v}{g}\right)^2} = \frac{2v}{g}$$ (we ignore the trivial solution that we didn't throw the stone in the first place, $v=0$)

So after half the time $t = \frac{v}{g}$ you reach a distance / height of s_{lin} = v^2/g. And going back, reversing the sign of velocity, you cover the same distance in the same time in the other direction.

To me personally the most intuitive image for this is not a stone thrown in the air, but a ship which tries to reverse: you start with a constant speed (thus motor needs no work, if we ignore water drag). Now, the motor provides the equivalent of the gravity force in this image, and tries to give speed to the boat in the exact opposite direction. Then it becomes (at least to me) intuitively clear, that it takes half the time to reduce the speed to 0. And then the same time to speed up again to the same speed, just opposite direction.

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Simply do the computation. With an initial velocity $v_0$ and an initial altitude 0, with an ascending $z$ axis: $$z(t)=-\frac{1}{2}\,gt^2+v_0t$$ Velocity is: $$\dot{z}(t)=-gt+v_0$$ The object reaches top altitude when $t=t_f$: $$\dot{z}(t_f)=-gt_f+v_0=0 \quad\Rightarrow\quad t_f=\frac{v_0}{g}$$ Now the altitude at instant $2t_f$ is : $$z(2t_f)=-\frac{1}{2}\,g(2t_f)^2+v_0t_f=0$$ so the object indeed took the same duration to reach top altitude and to return to its initial altitude.

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Maybe its easier to think about it this way: when you throw a ball straight up, the ball starts out with speed due to the force from your hand. As it ascends, it gradually loses its speed because of gravity. At the top, the ball has zero speed momentarily, and then it starts to fall downward - but again, now its starting out with zero speed. Then on its way down, due to gravity, it gradually regains speed and ends up with the same speed downwards, that you initially threw it upwards with, so there is symmetry in the ascent and the decent (in this idealized model).

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  • $\begingroup$ Can you please explain me, why velocity of projection = velocity of landing, in this situation. $\endgroup$ Aug 7 at 3:22
  • $\begingroup$ If you film the balls upward trajectory with your camera, then play the movie backwards but with normal playback speed, thats the exact path the ball will follow when it acutally falls down. (It can be shown formally that the two speeds have to be equal by appealing to the principle of conervation of energy. If your interested to learn more about that, you could read about concepts like potential- and kinetic energy and conservative forces.) $\endgroup$
    – user330563
    Aug 7 at 10:04

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