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Finding total flight time of a projectile. Please help!

  1. Nov 27, 2011 #1
    1. The problem statement, all variables and given/known data

    A baseball thrown vertically with a speed of 3m/s. Find the total time that the ball has been in flight when it had traveled 50cm.

    2. Relevant equations

    vf=vi + at
    d= 1/2at^2


    3. The attempt at a solution

    Found t.
    vf=vi +at
    vf=3
    vi=0
    a=~10
    Found t= .3 secs

    Found d
    d=1/2 + at^2
    d=4.5m

    Added 4.5m to .5m =5m

    Plugged in total distance 5m into d=1/2 + at^2
    Found t= 1

    Added 1+.3 to get total time of 1.3 secs.

    The answer in the manual is 0.398 sec.
    I am way off can someone please help. Thanks!!
     
  2. jcsd
  3. Nov 27, 2011 #2
    above equation is not always correct.
     
  4. Nov 27, 2011 #3
    grzz what do you mean?

    Is there another equation I should be using? or perhaps another approach. Im not sure where I went wrong.
     
  5. Nov 27, 2011 #4
    The equation that I quoted from your post is only correct to use if the initial speed is 0. This is not the case in your problem.
     
  6. Nov 27, 2011 #5
    I also tried using d=vit + 1/2at^2 plugging in 3m/s for vi but I also calculated a wrong answer for t. So I am not sure what equation to use.
     
  7. Nov 27, 2011 #6
    Use the constant acceleration equations. One in particular will allow you to find the final velocity at S = 0.5m, then you can use that to calculate other values e.g time.
     
  8. Nov 27, 2011 #7
    TaxOnFear-
    I tried using vf^2 =vi^2 + 2ad to find vf. I got sqroot of 19 for vf and then plugged it into vf=vi+ at to solve for t I got t=.135 sec... still the wrong answer according to the answer manual.
     
  9. Nov 27, 2011 #8
    You are assuming the projectile has not reached the peak point by this statement. Try breaking the problem down into two sections: one where the projectile is launched up, and the other where the projectile is falling down. Most of the numbers will look the same, but the time in the air will add up.
     
  10. Nov 27, 2011 #9
    I am really confused. Im starting to think there is an error in the answer manual. I have tried all motion equations possible. I assumed it reached peak and used vf=3m/s -- that didnt work. I tried vi=3m/-- that didnt work. I tried d= 1/2at^2 that didnt work either.

    Was my initial attempt at a solution (above) completely wrong?? Can someone tell me where I am going wrong. Ive tried looking at it from all angles..please help
     
  11. Nov 28, 2011 #10
    Better follow the advice of 'physicsvalk'.

    The 50cm may mean the total distance going up AND some distance going down.

    Hence try to find the MAXIMUM height it can travel. Of course in finding this you have to put the final velocity equal to 0.

    Then find the time to reach this max height and continue from there.
     
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