### Puzzle 2 : Which will take shorter Time?

“If you throw a stone up and considering real world situations like air drag ….. which will take lesser time —- reaching to the top or dropping from the top to where it started…”

Give reasons to support ur answer….

Sol>>> No one has given the correct answer and explanation yet… be the first one to answer it right…

Simple..

the time it takes to come down will be more as the air friction comes in effect

In case of going up the Forces of air friction and gravity work in same direction.. while in the opposite case when the stone is coming down the Forces work in different direction.. so the acceleration is less

simply put, in case when it’s thrown up the (de)accleration is

g+f

where f is (de)accleration due to Air friction

the case when it’s coming down the accleration is

g-f

Hence the answer !!

My solution hidden !!

Waise Arindam almost all of us were IIT-JEE aspirants ðŸ˜€

Getting to the top will take less time. One can throw a stone considerably quicker than 9.8 m/s/s.

Robert the answer is wrong!

But we better wait for the author for revealing the correct answer

Hiding the solution anyways ðŸ™‚

arre dont hide the answers that are wrong! others wont make the same mistake naa ðŸ™‚

what is the correct ans?

ok i have un-hidden the answers submitted.. now wait for the author to give out correct answer.. or speculate ðŸ˜›

Well Abhi u missed something and I would ask u to think about this….

Air drag is directly proportional to velocity and also to the size of the object thrown. I hope u have heard of how dimples in a golf ball results in the ball covering larger distance and reaching more height…. if not go to Resnick Hallidey essay on aerodynamics..

You cannot replace air drag with a constant acceleration. Now thing on terms of conservation laws and try to find the answer.

The time will be the same going up as going down. When the stone is thrown up the initial kinetic energy is gradually turned into potential energy as it climbs higher untill it is all converted to potential energy when it then stops and falls back to earth. The potential energy is then converted back into kinetic energy which means when it reaches the ground the potential energy is converted back to the same kinetic energy as it started with. kinetic energy is related to speed which is then related to the time it takes to travel the two routes. The air resitance is the same in both directions, but will vary with speed. On the up throw the airresistance will start off high and drop to zero as the stone reaches the top of its flight, then it will reverse its path with little air resistance to start with then reaching maximum as it reaches the start point. It will mean some of the energy is disapated as heat but this is the same going up as going down.

Ya Nigel u are not correct. Energy is lost due to friction so u cannot say that the whole of kinetic energy is converted to potential energy. So it will reach lower height then what was to be expected. Now as the frictional force is directly proportional to speed, so it will start from are greater drag force which will come down to zero at the top and when the object starts falling the drag force will not reach the value that it attained at the start. So integrating in the two paths, the amount of retardation while going up is more then the amount of retardation while coming down.

Now one sees that at the time of going up the force acting downwards is g + drag force up, and the force while coming down is g – drag force up. But as the drag force is dependent on initial velocity and shape of the object thrown, so for a smaller body , it will depend on weather v/(g+drag-up) is greater than v/sqrt( drag-up * (g -drag-down))/ which is true for a smaller body. So the time taken to come down for a smaller body is less then while it goes up. But when thrown at a larger speed, then things can change.

Also how one throws also defines the time taken. If one throws a ball swinging, then there will be a pressure difference created between the top of the ball and the bottom, which can give the object lift( as what is seen in case of golf balls whith dimples on their surface). In these cases it takes more time to reach to the top than coming down.

But at high speeds things can change.

The time it takes to come down from the top will be more.

Reasons:

we all know this equation v*v = u*u + 2*a*s

my first job is to find the mod of acceleration in each case.

When going up equation will be u*u = 2*|a|*s

while coming down , we know that because the frictional force is lessening its total energy, and some of its energy will be lost as part of frictional energy or heat. As the potential energy will be the same at the time of throwing up and after coming back, so it must be the kinetic energy that will be converted in heat and dissipated. So, equation will be (u-i)*(u-i) = 2*|b|*s, where i is a +ve number.

Hence we have b < a or b = a-j.

now use s = ut+(1/2)at*t

in case 1, while going up, we have s = u*t+ (1/2)at*t

in case 2, while coming down, we have s = (u-i)t’ + (1/2)b*(t’)(t’)

i.e. s = (u-i)t’ + (1/2)(a-j)*(t’)(t’)

clearly t’ must be greater than t , for above 2 equations to satisfy. Hence time in coming back will be more.

Hemant is correct, though I believe there is a way to imagine it for those who don’t understand the equations. Suppose that instead of a stone, you throw a balloon in the air (and lets suppose we are indoors so there is negligable wind). It will not go very high, but the original impetus will probably mean it stops a meter or so higher than where it started and will reach this point fairly quickly. It will then float back down very slowly indeed, thus proving that it will take longer to come back down. Like I say, I believe this will hold for any object.

Ignore my last message. The balloon has a low terminal velocity so it doesn’t compare. The time taken is the same. As Nigel points out…

How about this: The stone takes longer to drop back down. Reason:

(mv^2)/2 = mgh

The time taken for the stone to get to h is the average speed of the stone multiplied by the distance, h, itself.

When the stone gets to h, its kinetic energy has been transferred to potential (gravitaional) energy – but not entirely: some energy is lost due to resistance.

So 1/2mv^2 > mgh (where v = release velocity)

Similarly, when the stone falls, _most_ of the potential energy is transferred to kinetic energy, but not all of it: again, due to air resistance

So mgh > 1/2mz^2 ( where z = return velocity)

So the speed of the stone when you catch it, is less than the speed when you let go. Therefore, the average speed of the stone on the way up, is greater than the average speed on the way down, and hence, it takes longer to return.

Will I have given the reply as comment no 10. Please go through it, if the solution does not seem satisfactory then do comment, In ur solution u have totally neglected size.

Well, isn’t this a question of energy conversion? Assuming the ideal case without any air drag, all the kinetic energy imparted during the upward throw gets converted into potential energy (100% efficiency) and on the way back, all of that gets converted back into kinetic energy. However, in the realistic case, part of the imparted kinetic energy is lost as heat. So, when the ball reaches the top, only a portion of the initial energy is stored in it. With this lesser energy, it is going to get longer to come back down to where it was thrown from. So, I think the downward trip will take longer than the upward trip.

The time is same for both coming up and down.

Hi!,

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The time taken for the both will be the same

for up ward journey

Vup^2 – Uup^2= -2gh here Vup = 0

therefore

Uup^2= 2gh —-> h = Uup^2/2g ——(1)

also

Uup = Vup + gTup since Vup = 0

Uup = gTup ——– (2)

from (1) & (2)

h= (gTup)^2/2g

on simplyfying

Tup^2=2h/g———————(A)

—————————————————————————————

for downward journey

Vdo^2 – Udo^2 = 2gh here Udo = 0

so

Vdo^2 = 2gh ——> h = Vdo^2/2g ——(3)

also

Vdo = Udo + gTdo Udo = 0

so Vdo = gTdo —————–(4)

from (3) & (4)

h =(gTdo)^2/2g

so Tdo^2 = 2h/g —————(B)

from (A) & (B)

Tup^2 = Tdo^2 hence both times are equal. As for as drag

it is same for both upward and downward motion.