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Time that an object is in the air

  1. Nov 30, 2005 #1
    In a question, it asks for the time that an object is in the air; at first I thought it meant the time at max height, but I was told it should be the time at max height divided by 2 because the object goes up and down. But if the object goes up and down, wouldn't you mult by 2?
     
    Last edited: Nov 30, 2005
  2. jcsd
  3. Nov 30, 2005 #2
    The time it takes to reach max height x 2 would give you the total time it was in the air.

    Either your reading a typo or I'm reading the question wrong.
     
  4. Nov 30, 2005 #3
    Here's the question:
    An object fired straight upward with an initial velocity of 1.0 x 10^2 m/s is subjected to the effects of gravitation.
    In a previous question, I solved for max height, which was 5.1 x 10^2 m. (I'm not sure if this is related to the next question though).
    The time the object is in the air is
    a. 4.5 s
    b. 9.0 s
    c. 20 s
    d. 10 s

    I keep on getting d. 10s

    vi(velocity initial)= 100m/s vf(final)= 0 a(acceleration)= -9.81m/s^2 displacement=not needed t(time)= ?

    a = (vf - vi) / t
    t = (vf - vi) / a
    t = (0 - 100m/s) / (-9.81m/s^2)
    t = (-100m/s) / (-9.81m/s^2)
    t = 10s but the correct ans should be a. 4.5s :frown:
     
    Last edited: Nov 30, 2005
  5. Nov 30, 2005 #4

    kp

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    hmm...sounds strange too me. 4.5s total time for a projectile at shot 100m/s?
     
  6. Nov 30, 2005 #5
    well, i could be wrong about total time, the question maybe asking for smthg else... I geuss I'll check if the ans key is wrong tomorrow... :wink:
     
  7. Nov 30, 2005 #6

    kp

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    yeah I get H = 510.20 meters, total time 20.4seconds, time at H(max) = 10.2 seconds
     
  8. Nov 30, 2005 #7
    and even if you divide by 2 for whatever reason, you still wouldn't get 4.5s, so...

    well, thanks!
     
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