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Free Fall

  1. Jan 2, 2012 #1
    Hi everyone, Im doing a lab report about free fall, but I have to find the terminal velocity for the balloon, but I dont know if I can use the formula velocity = Distance / Time because I have 10 measures from free fall ( 0.2 m - 2 m) , and I have the time that it took to hit the ground.
    For example;
    0.2 cm reached the ground on 0.41 seconds.

    Using Velocity = Distance / Time
    0.2 / 0.41
    V= 0.4878

    *I also have to graph the velocity from the balloon. So is why I thinking of doing Velocity = Distance / Time and with the graph the last measurement I will get the final velocity but I not sure..
     
  2. jcsd
  3. Jan 2, 2012 #2
    No you cannot use v=d/t because that eqn is valid only when acceleration is 0
    Do you know any equation for distance when there is some acceleration ?
     
  4. Jan 2, 2012 #3
    While it is against the rules of this forum to tell you how to go about this experiment, think about the shape a graph of times versus heights and what can be deduced from the plot and your equation.
     
  5. Jan 2, 2012 #4

    BruceW

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    well, total distance/total time = average velocity. Now you have to think: does the average velocity of the balloon approximately equal the terminal velocity of the balloon? And you'll need to justify your answer in the lab report.
     
  6. Jan 2, 2012 #5
    well I know this vt = √(m.g/k) but this is for find Terminal Velocity
     
  7. Jan 2, 2012 #6
    Hint: Look at the shape of the graph you plot.
     
  8. Jan 2, 2012 #7
    Thanks, but can I do a graph ?? because I read on my notes that If a draw graph of the falling velocity of the balloon and see how it flattens as the acceleration gets close to zero this is the terminal velocity ... is why I was thinking of doing velocity = distance / time , but my first time don't starts on 0s ..
     
  9. Jan 2, 2012 #8
    The first graph Height vs Time?
     
  10. Jan 2, 2012 #9
    A graph of heights versus times necessitates numerous drops from different heights with respective timings. By its shape, what would such a graph demonstrate?
     
  11. Jan 2, 2012 #10

    BruceW

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    Yes, the first graph you should draw is height vs time taken to reach ground because these are the direct measurements you took.
     
  12. Jan 2, 2012 #11
    Yeah I doa lot of drops, and the graph is a straight line I got the gradient but I don't think that is necessary.
     
  13. Jan 2, 2012 #12
    I already done with that graph, now I have to draw a graph about ballon vs velocity.
     
  14. Jan 2, 2012 #13

    BruceW

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    If you purely use the results you've taken, then you can only get the average velocity for each drop (not instantaneous velocity).
     
  15. Jan 2, 2012 #14
    OK so can I use Distance / Time= average velocity or do I have to use total Distance / Total time ?
     
  16. Jan 2, 2012 #15

    BruceW

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    for each drop, you've recorded height and time. These are total distance and total time for each drop. So height/time will give you average velocity for a particular drop.
     
  17. Jan 2, 2012 #16
    Sorry for bothering BruceW but for example for 0.2m it took 0.41 seconds to hit the ground..
    So height/time 0.2 / 0.41
    Average velocity= 0.4878 for 0.2 m
     
  18. Jan 2, 2012 #17

    BruceW

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    exactly. That is the average velocity for that drop.
     
  19. Jan 2, 2012 #18
    Thanks! So now with the average velocity for each drop I can graph the velocity vs balloon. But the terminal velocity is the last drop measurement? In this case is 2 m am I right? Because in my notes I saw a graph of velocity vs an object and the last measurement is the terminal velocity in that graph..
     
  20. Jan 2, 2012 #19

    BruceW

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    generally, no. The average velocity is the average velocity over the entire drop. So it would only be equal to the terminal velocity if the balloon was travelling at terminal velocity the entire time.

    The 2m drop would give an average velocity which is closer to the terminal velocity than the other drops. Can you think why this is?
     
  21. Jan 2, 2012 #20
    because acceleration goes close to zero :/ I dont know :/
     
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