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Free fall object

  1. Aug 31, 2010 #1
    1. A ball is thrown vertically upward with an initial speed of 10m/s from thetop of a building, at a height of 25m from the ground. Assume free fall conditions.
    a. To what maximum height above the ground does the ball rise?
    b. With what velocity does the ball strike the ground?
    c. How long does the ball take to reach the ground?


    2. V=v0+at
    y-y0=v0*t+1/2at^2




    3. For the first part (a) i used v=v0+at and plug in my knowns
    v=0, v0= 10m/s, and i solved for t which came out to be t=1.0 sec. With that i can find my maximum which I plug into the equation y-25=10(1)+(1/2)(-9.8)(1^2) and got the answer 30.1m
    and for part b and c, I have no clue how to solve it if anyone can help me with a simple problem that be great
    thanks for the help!
     
  2. jcsd
  3. Aug 31, 2010 #2

    diazona

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    Part (b) is asking for a velocity. What do you know that might allow you to calculate velocity?
     
  4. Aug 31, 2010 #3
    v^2=v(int)^2+(2)(delta y)...?
     
  5. Aug 31, 2010 #4
    i found v^2=10^2+2(-9.8)(-25)
    did the math and v=24.29m/s when it hits the ground
     
  6. Aug 31, 2010 #5

    diazona

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    OK, now part (c) asks you for a time. What do you know that could allow you to calculate time?
     
  7. Aug 31, 2010 #6
    the first equation v=v0+(-9.8)t solve for t which i have (v-v0)/(-9.8)=t
    and i got -1.45s which is totally wrong since time cant be negative, I am not surewhere i have done wrong...
     
  8. Aug 31, 2010 #7

    diazona

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    Can I see your work? What values did you plug in to the equation? (other than -9.8 m/s² of course)

    By the way, time actually can be negative sometimes. Whenever you do a physics problem involving time, you have to pick some event to be t=0. If something happens before that event, it will have a negative time coordinate. It's a lot like space coordinates (x, y, z); you have to pick a point to be, say, x=0, and something that is located to the left of that point will have a negative x coordinate.

    In this case I think you're right, though, that the time should be positive.
     
  9. Sep 1, 2010 #8
    i have plugged in
    (24.29-10)/-9.8=-1.45
     
  10. Sep 1, 2010 #9

    diazona

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    Which direction is positive, up or down?

    Once you figure that out, make sure the numbers you plug in have the correct signs.
     
  11. Sep 1, 2010 #10
    well it should be (24.29+10)/9.8 but I am not getting the concept, I know that 24.29 is the velocity when it hits the ground and 10 is the initial velocity so they have to be positive but I dont know about gravity since gravity is always pull downward so it has to be negative right...? if you can explain it to me that be great
     
  12. Sep 1, 2010 #11

    diazona

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    OK, so downward is negative, right?

    Remember that velocity is a vector: it has a magnitude and a direction. What is the direction of the 10 m/s? What is the direction of the 24.29 m/s?
     
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