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Throwing ball in air and final velocity.

  1. Dec 5, 2011 #1
    1. The problem statement, all variables and given/known data

    You throw a ball straight up @ 22m/s and you catch it 3.6 seconds later. How fast was it going when you caught it?



    2. Relevant equations


    [tex]v_f=v_i+at[/tex]

    3. The attempt at a solution

    I found the right solution. But to me, it seems it should be more complex then this.
    The ball goes 22m/s second and gravity had its opposition at 9.8 m/s squared going up but going down 22m/s has nothing to do with it falling. So why wouldn't this be 2 equations, one for going up with v_i of 22 and the other for going down with v_i at 0.

    The answer of course is 13.28m/s with equation above.

    Perhaps this has something to do with a quadratic equation, but I don't know how to see this with the correct perspective.


    I did look at this though and am just unsure how to really view this
    http://www.wolframalpha.com/input/?i=-9.8x^2+%2B+22x
     
  2. jcsd
  3. Dec 5, 2011 #2
    The initial velocity determines the time for which the object is in air. You must got the answer with a negative sign. So, it means that the velocity has changed direction after becoming zero at some point. Kinematical equations also give the data after velocity has reversed the direction. So no need to to use 2 eqn the -ve sign is the key thing.
     
  4. Dec 5, 2011 #3
    So is the ball in the air longer when it is traveling downward and shorter when traveling upward?
     
  5. Dec 5, 2011 #4

    NascentOxygen

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    Staff: Mentor

    v = u + at

    Going up, until standstill: 0 = -22 + gtr ..... (eqn. 1)
    so we can find t, if we needed to.

    Going down, until it reaches 22m/sec: 22 = 0 + gtf ..... (eqn. 2)

    Compare eqns (1) and (2), you see they are identical, so tr = tf

    In other words, it takes as long to descend as to ascend. Motion is affected by gravity while rising, and while falling, identically.
     
  6. Dec 5, 2011 #5

    But it doesn't equal 22m/s when he catches it. it equaled 13.28m/s. So did wouldnt he of had to catch it in a higher position then when he threw it?
     
  7. Dec 6, 2011 #6

    NascentOxygen

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    Staff: Mentor

    Yes. Apparently he jumped 15 metres into the air to catch it early. (Use s=ut+0.5at2 to determine the ball's location after 3.6 secs.)

    The ball took 2.245 secs to become momentarily stationary, so it would take the same length of time to descend to the height it was launched from.

    This seems like a trick question.
     
  8. Dec 6, 2011 #7

    NascentOxygen

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    Staff: Mentor

  9. Dec 6, 2011 #8
    No as nacsent oxygen has demonstrated. I just said that if you throw it up with larger velocity, it will take longer to change direction.
    And if you add the 2 eqn you were mentioning you would get the basic eqn only.
    [itex]0=v_{i}+at_{1}[/itex]
    [itex]v_{f}=0+at_{2}[/itex] adding
    [itex]v_{f}=v_{i}+at[/itex] as t1+t2=t
     
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