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Okay so I am studying 2d kinematics. And I am woundering, when an object reaches its highest point its vy=0. So when an object falls back down from its highest point, does it fall back down at the same time it took to go up? Since gravity acts on the ball downward, the ball should be traveling faster so it should hit the ground at a much faster rate. But my friend says, the time is the same for when it goes up and reaches Vy=0 and when it comes down and hits the ground. Whos right...and why? Thanks in advance

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  • $\begingroup$ If you are ignoring every force but gravity your friend is correct. All you need to do is calculate it yourself using the kinematics equations using some initial velocity, you can then know the max height and then the time it will take to hit the ground again. If this were not true projectiles wouldn't take parabolic paths through the air. $\endgroup$ – Triatticus Apr 22 '18 at 20:08
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The uniform accelerate moving is symmetrical.

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$V_C^2=V_A^2+2 g \Delta y$ $\space \space \And \ \space$ $\Delta y=0$ $\implies$ $V_C=-V_A$

$V_B=V_A -g \Delta t_{AB}$ $\space\And \; \ \ V_B=0$ $\implies$ $ \Delta t_{AB}= \frac{V_A}{g}$

$ V_C=V_B -g \Delta t_{BC}$ $\space\And \; \ \ V_B=0$ $\implies$ $ \Delta t_{BC}= -\frac{V_C}{g} \implies \Delta t_{BC}=\frac{V_A}{g} $

$\Delta t_{AB}=\Delta t_{BC}$

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When the ball is travelling up, the gravity acts against it, so the equation that works is. $$\frac{\mathrm{d}v}{\mathrm{d}t}=-g \implies \int_{u}^{0}\,\mathrm{d}v=\int_{0}^{t_{up}}-g\,\mathrm{d}t \implies-u=-gt_{up} \implies t_{up}=\frac{u}{g}$$ When the ball is travelling down from the topmost zero velocity position, then it will return to the earth with the same final velocity $u$ owing to conservation of energy.

Hence, the equation for the downward motion is $$\frac{\mathrm{d}v}{\mathrm{d}t}=g \implies \int_{0}^{u}\,\mathrm{d}v=\int_{0}^{t_{dn}}g\,\mathrm{d}t \implies u=gt_{dn} \implies t_{dn}=\frac{u}{g}$$

Thus $$t_{up}=t_{dn}$$

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