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Why is the time of ascent less than the time of descent in a projectile motion.

I understand that while going up the air resistance and gravity act downwards and while coming down gravity is downwards and air resistance upwards but I still dont get it.

Is there another explanation or perhaps a mathematical proof to this ?

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    $\begingroup$ How are you accounting for the air resistance? Look closely at that term. $\endgroup$ – Jon Custer Mar 17 '16 at 14:34
  • $\begingroup$ Related: physics.stackexchange.com/q/40778/2451 , physics.stackexchange.com/q/96190/2451 and links therein. $\endgroup$ – Qmechanic Mar 17 '16 at 14:55
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    $\begingroup$ You say in your question that the forces on the projectile are different in the up and down part of the trajectory, because gravity only ever acts downwards while air resistance acts opposite to the direction of motion. Given that the forces are different in the two parts of the trajectory it would be astonishing if the up and down parts were identical. $\endgroup$ – John Rennie Mar 17 '16 at 16:58
  • $\begingroup$ one can see a detailed analysis of motion of a projectile with air resistance and it happens that the path of the projectile in rising period and the path in the descent does not remain the same. see >farside.ph.utexas.edu/teaching/336k/Newtonhtml/node29.html $\endgroup$ – drvrm Mar 18 '16 at 4:18
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For simplicity assume the resistance to be be constant F. During ascent mg and F both oppose motion of the body. During descent F opposes motion while gravity supports it. Thus the rate of decrease in velocity during ascent is greater than the rate of increase in the velocity during descent.

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The neatest answer I know to this uses energy rather than force....

The projectile passes any height $h$ twice, on the way up and on the way down, and in both cases it has potential energy $mgh$.

Between these two occurrences it loses some energy - maybe a little, maybe a lot, we don't have to specify any particular model - due to air resistance, so its total energy, which is $mgh+{1 \over 2} mv^2$, must be smaller on the way down than on the way up.

But as $mgh$ is the same going up and coming down, ${1 \over 2} mv^2$ must be smaller on the way down, and that means $v$ must be smaller.

So at every point on the trajectory, the speed is greater on the way up than on the way down. So the upward journey is quicker than the downward journey.

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Let resistance of air be $R =ma$. The resistance always act in the opposite direction of motion. So during ascent it acts downward and gravity also acts downward so total acceleration add to give $a= g+R/m$.

But during descent the resistance acts upward and gravity acts downward so total acceleration is $a= g - R/m$.

Now it is clear that during descent acceleration is less and we know that acceleration is inversely proportional to time because $a = v/t$. So time of descent is always greater than time of ascent.

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 Here is the mathematical explanation with simple equations of motion. Any further suggestions is welcomed. If we take into account the air resistance, then time of acsent is less than the time of descent.

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