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Free fall acceleration please help

  1. Sep 10, 2007 #1
    1) A lead ball is dropped into a lake from a diving board 5.12 m above the water. It hits the water with a certain velocity and then sinks to the bottom with this same constant velocity. It reaches the bottom 4.72 s after it is dropped. (Assume the positive direction is upward.) How deep is the lake?
    For this question I dentified the velocity at which the ball hits the water and that is 10.02 m/s. To get that i used the formula v^2=Vi^2+2ax. I used vi=0, 9.8=a x=5.12. I do not know what to do know because i keep getting the wrong answer and its not off by like like 1 or 2 its off by 10.

    2) An object falls from height h from rest and travels 0.46h in the last 1.00 s.
    Find the time of its fall.
    Find the height of its fall.
    For this one I believe i should use the formula x=vt-.5at^2. I know believe time is t-1 and the hieght is h-.46. Could you please help me understand what i should do next becuase i am clueless how to put this information into a formula.
    Thank you for the help
     
  2. jcsd
  3. Sep 10, 2007 #2

    Doc Al

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    So far, so good.
    Start by figuring out how long it takes for the ball to hit the water. Then figure out how much time it spends in the water.
     
  4. Sep 10, 2007 #3

    Doc Al

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    That will work, using down as positive and thus D = 0.5gt^2. But you need to apply it twice: once for the distance h-.46h and once for the distance h.

    Correction: The first position given is .46h, not .46; so the distance fallen is h-.46h.
     
    Last edited: Sep 10, 2007
  5. Sep 10, 2007 #4
    i dont understand what your saying by using twice how would i put that into a formula and how would i solve for t and h because there would be 2 varibles in that equation you gave me
     
  6. Sep 10, 2007 #5
    btw thanks for your help with the first problem it made it a lot easier for me to solve
     
  7. Sep 10, 2007 #6

    Doc Al

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    Since you have two variables you need two equations. Write one equation for the distance h-.46h & time t-1; write another equation for the distance h and time t. You'll need to solve them together to find h & t.

    (See my correction to post #3.)
     
    Last edited: Sep 10, 2007
  8. Sep 10, 2007 #7
    okay i do that and i come up with the equations h=.5*9.8*t^2 and h-.46h=.5*9.8*(t-1)^2 so now what do i do with that is there a way to set them equal and solve for t because i cant figure out how to do that
     
  9. Sep 11, 2007 #8

    Doc Al

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    One way to solve these simultaneous equations is to substitute one into the other. Replace "h" in the second equation with .5*9.8*t^2 from the first equation. Then you can solve the quadratic equation for t. (And then plug in to get h.)
     
  10. Sep 11, 2007 #9
    Thanks Doc Al for all your help. I finally got the answer. Thanks so much.
     
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