Question

In real life condition, time of ascent of a freely falling body under gravity is equal to time of descent of that body or not? Please define.

Answer 1

but in real life condition.

Answer 2

but yaar it should also depend on the type of body,,, and other parameters such as weight, size, density, air resistance etc.
but in general there will less time of ascent then time of descent.

Answer 3

Oh! thanx.

Answer 4

looks like you got help already :)
but yes gravity pulls bodies toward earth so descent would logically take less time

Answer 5

but viendra said that ascent will take less time & you are saying vice versa.please tell me why.

Answer 6

and answer of this question is also as viendra's.

Answer 7

oh i misread their answer...oops well im not a physics expert
maybe i should ask what you mean by ascent of a falling body?

Answer 8

but please tell me why this happen?

Answer 9

Why this phenomenon of falling bodies happens and no problem for last statement.

Answer 10

To my knowledge this is called vertical motion and the time is the same

Answer 11

but this not the answer.

Answer 12

the answer is as viendra said in first comment

Answer 13

Seriously I cant remember

Answer 14

but I will try to get the answer by tomorrow

Answer 15

later today

Answer 16

no problem.

Answer 17

I think time of descent will be greater, because, by symmetry, the x components of the resistive force are equal, while in y components the acceleration's modulus(in case of ascent) is more than that in case of descent so, time (y-axis) = velocity/acceleration where \(velocity = |v_{final} - v_{initial}|\) during its trajectory in the y axis. So I am with @viendra here.

Answer 18

That's quite a simplification, it depends a lot on the initial speed, body properties, and so on. Doing a physicist's approximation, I would agree that, in general, the time of descent is greater.

Answer 19

But what is the meaning of x components of resistive forces there?

Answer 20

That is me saying that I am disconsidering x-moviment. Draw a simplified force diagram to see that, for ascent, the air resists the moviment of the body with the acceleration being in the same direction as the resistance, so to speak. It's the other way around for descent. That's basically my hypothesis.

Answer 21

Oh! I understood now.Thanx a lot!

Answer 22

Or, if you prefer:
air and g both act opposite to movement of a body upwards: \(\Large t_{ascent} = \frac{v_0}{g+air} \)

air tries to retard the body and g accelarates returning to ground. \(\Large t_{descent} = \frac{ v_0}{ g - air } \)

Answer 23

Where air is the desacceleration caused by air resistance.

Answer 24

oh I see

Answer 25

thanx!

Answer 26

No problem :-)