A stone vertically thrown up with velocity v takes t seconds to reach back without cosidering air resistance. Now due to the air resistance will this time increase, decrease or remain same ? Pls justify your answer! I don't know the answer.

Probably air resistance do not effect the velocity of  a particle !!!! since it has its definite value then also it is so small that it could be neglected !!! so the time 't' will remain same !!! CORRECT ME EVERYONE IF I'AM WRONG!!!!

Suppose it is a  round stone whose density is far more than air. I felt that it will go to a lesser height due to air resistance and so total traverse time may not change. We are discussing the effect if it is not neglected. It is true that effect is negligible. Still I am eager to know what it is?

Pls continue discussion!!!

i guess it will take less time bcoz due to air resistance it will travel less distance and thus will take less time to come back.........

While going up the forces opposing motion are mg+f  instead of mg, so time to reach max. height will be less(ofcourse height will be less).

While coming down force mg supports motion and f opposes, (mg-f) so it will be slower.

Net result could be that there will be no change in total time.

After seeing the varying opinion of everyone, I feel that question itself was wrongly placed, in the sense that there can't be a definite answer as it depends on various factors such as initial velocity,stones density, its shape, terminal velocity etc.

Dude, I think total time will not change, but reason is different.

We know that period of a simple pendulum does not change when amplitude comes down due to air friction!!!

So here also time may not change if it is a stone.

cant say so easily...

y dont u juss calculate... problem is tha whn there is an additional force opposing motion distance travelled in two cases is different.... and while going up and coming down acceleraton of body changes in case of air resistance... a is acc due to friction by air

like let initial vel =u

final vel=0.. acc =g => time taken = u/g...

total time = 2u/g

now

2nd case: whle going up ... 0=u-(g+a)t  => t= u/(g+a)

and 0= u^2 -2(g+a)s =>s=u^2/2(g+a)

now while going down

s is same.... but acceleration is g-a

=> s= 1/2 (g-a)t^2...(putting value of s)

=> t^2 =u^2/(g+a)(g-a)... => t= root of u^2/(g+a)(g-a)...

now total time=

u/g+a +u/root(g+a)(g-a).... =/= 2u/g

we thus have conditions depending upon magnitude of a....

cheers!!! :)

@saurabh_reincarnated

Thanks for the effort. I think it is a complex problem. Your 'a' is not constant. It is a function of velocity.

As per this formula it appears that time is decreasing.

TIME OF ASCENT..

LETZ TAKE THE NET FORCE ON THE BODY...WHEN IT IS IN ITS ASCENT..

NET FORCE=mg+F_air

mg'=mg+F_air

g'=g+F_air/m

T_a=u/g=(root2g'h)/(g+f/m)

    =root(2(g+f/m)h/(g+f/m)²)  (by substituting g'=g+f/m)

    =root(2h/g+f/m)

similarly.....T_d=root(2h/g-f/m)

so t_a less than t_d.

RATE ME IF MY ANS IS SATISFACTORY....