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Time and air resistance question; throwing a ball upward.

  1. Sep 10, 2012 #1
    1. The problem statement, all variables and given/known data

    "If a ball is thrown upward in presence of air resistance, would you expect the time during which it took to rise to be longer or shorter than the time it takes for it to fall. (use the principle of exaggeration)"


    2. Relevant equations



    3. The attempt at a solution

    I said it depends on the speed at which the ball is thrown. If you throw the ball at a speed that is less than its terminal velocity then the ball should take almost exactly the same time to rise as it does to fall. On the other hand, if the ball is thrown faster than its terminal velocity speed, the ball will take a longer time to fall due to it reaching terminal velocity on the way down.

    The books answer just says that the ball will take longer to fall. I obviously don't think that is true for all situations, but that is why I am here to check. Is my answer correct or am I wrong and the ball will ALWAYS take longer to fall? If I am wrong, please explain how that is possible. Thank you!
     
  2. jcsd
  3. Sep 10, 2012 #2
    I'm guessing since air resistance is dependent on velocity that when throwing the ball upward since both air resistance and gravity are applying force in the same direction the ball will more quickly come to a stop however since when the ball is dropping the force due to air resistance is opposing gravity the ball will take longer to travel the same distance back down.
    Just a thought though. As far as terminal velocity goes I don't think that matters because the ball upward would immediately not be traveling at terminal velocity and then going back down it would obviously take time to get back up to that speed.
     
  4. Sep 10, 2012 #3
    The force acting on the object on upward motion is the net force of 2 forces against it, friction plus gravity.
    On way down, the net force is gravity minus air resistance.
     
  5. Sep 10, 2012 #4

    PeterO

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    Suppose the object was a party balloon, which was thrown up at 10 times the speed of sound.
    It will stop going up pretty quickly, but still have travelled to quite a large height.
    It will then drift down at its small terminal velocity, taking an enormous amount of time to reach the ground.
     
  6. Sep 10, 2012 #5
    I appreciate all of the replies but nobody has taken my answer into consideration completely. I understand that when an object like a party balloon is thrown up faster than its terminal velocity it will go up faster than it descends. I have demonstrated that fact in my answer. The part that I need confirmation on is my logic where a balloon or ball is thrown up at a speed half its terminal velocity, it will go up and then when falling to the same level it was thrown it will only accelerate to the same speed it was thrown (half its terminal velocity) and with that being said, it will take the exact same amount of time going up as it did going down. The air drag will calculate to the same amount of friction because it never reaches terminal velocity and therefore accelerates equally up as it did down.

    Thank you.
     
  7. Sep 10, 2012 #6
    during upward motion the object has to overcome two forces i,e..friction and gravity due to which the object stops quickly but during its downward journey one force will help the object to reach quickly but the other force which is air resistance will oppose the object and delayed the time for reaching the point quickly
     
  8. Sep 10, 2012 #7

    ehild

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    ,

    Your logic is not correct. When thrown upward, the speed will decrease anyway either the initial speed exceeded the terminal speed or not. When dropped, the speed will increase, but never reaches the initial speed, as the air drag is not a conservative force, it decreases the energy of the ball all along its travel.
    Have you studied differential equations? You could solve the equation of motion h(t) (h is height) of the ball, and then you really could see how the raise time is related to the time of fall.

    ehild
     
  9. Sep 10, 2012 #8

    PeterO

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    My suggestion came out of following the instruction "use the principle of exaggeration".

    Your answer relating to projection at a speed equal to half its terminal velocity does not take that suggestion, which is why I did not reference it.
     
  10. Sep 10, 2012 #9

    ehild

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    The differential equation for the velocity is
    [tex]\frac{dv}{dt}=-g-kv[/tex]. In terms of the terminal velocity, [tex]\frac{dv}{dt}=-k(V_T+v)[/tex]
    The solution:[tex] v(t)=(v_0+V_T)e^{-kt}-V_T[/tex].
    The time of rise is [tex]t_{up}=\frac{1}{k} \ln\left(\frac{v_0+V_T}{V_T}\right)[/tex].
    The height y is the integral of v(t):
    [tex]y(t)=\frac{V_0+V_T}{k}(1-e^{-kt})-V_Tt[/tex].

    If the time of fall would be the same as tup, the ball would fall back to the ground in 2tup time. Substituting 2tup ln(3/2)) for t in y(t), if you get a positive height, longer time is needed to fall to the ground than to rise to the apex.

    ehild
     
  11. Sep 10, 2012 #10
    Thanks, I have not gotten that far in my physics classes to have even seen those equations yet. The more I learn about friction and seemingly simple things like a ball being tossed in the air, the more I realize that I have a lot of learning to do. I love it though and physics will be a lifelong love of mine. :)
     
  12. Sep 11, 2012 #11
    I hope you see (based on what multiple people here said) and even without dealing with the differential equations that you can intuitively know the answer just by recognizing that the direction of force due to air resistance is the same as gravity when the ball travels upward and in the opposite direction as gravity when traveling downward.

    And trust me, the further you get in your physics classes, the more you'll realize how little you and everyone else really knows...
     
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