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Flower Pot Falling Past a Window

  1. Sep 7, 2004 #1
    As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length of your window is L_w. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity g.

    Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).

    now i need to find the height at which the pot was dropped. im using the equation V_f^2 = v_i^2 + 2a(x_f - x_i)

    i figured out the final velocity to be L_w/t + g*t/2

    and the initial velocity is 0 because it is dropped but when i enter that the height is (L_w/t + g*t/2) / 2*g , i get an incorrect answer. then i thought of doing the sqrt(L_w/t + g*t/2) / 2*g and it was still wrong. can someone explain what i am doing wrong?
     
  2. jcsd
  3. Sep 7, 2004 #2

    Doc Al

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    Staff: Mentor

    Check this one over.
     
  4. Sep 7, 2004 #3
    what is wrong with it? there was a part of the problem where it asks to figure out the final velocity the instant it passes the bottom of the window and that was correct.
     
  5. Sep 7, 2004 #4
    Listen to Doc Al. There is something wrong with the equation you gave. Rework the derivation of the equation and you will see why.
     
  6. Sep 7, 2004 #5

    krab

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    Where did the square root come from? Should be squared. Always check your formulas if they are dimensionally correct. Yours are not. Your first one has units of velocity/acceleration, which is Time, not distance. Your second one has units of [itex]\sqrt{v}/a[/itex]. This is certainly not distance either.
     
  7. Sep 7, 2004 #6
    i see now, i forgot to square the final velocity
    thanks all for your help!
     
  8. Sep 7, 2004 #7

    Doc Al

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    Since I found an error in your expression for the final velocity, I didn't look at this part too carefully. As krab explained, always check the dimensions/units of your answer to see if they make sense.

    But leaving aside the fact that you have an incorrect expression for the final velocity, you are plugging the wrong thing into the kinematic equation you gave. You need to input the final velocity squared.


    Edit: Your expression for final velocity is correct--my mistake.
     
    Last edited: Sep 7, 2004
  9. Sep 7, 2004 #8

    Doc Al

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    oops... my bad

    Guess what? There was nothing wrong with it. I wasn't paying attention--I thought you were calculating the initial velocity (at the top of the window).

    My apologies! :blushing: (Back to sleep for me.)
     
  10. Sep 7, 2004 #9
    actually i could use your help for one more part of the problem. it is asking for the final velocity as the pot hits the ground. it introduced a new velocity v_b as the velocity at the instant the pot is at the bottom of the window, thus, i am able to use that velocity v_b as the initial velocity and h_b is the distance from the bottom of the window to the ground.
    so v_b is just the final velocity from the last problem : L_w/t + g*t/2

    so the velocity when the pot hits the ground should be

    v_b^2 + 2*g(h_b) but this was wrong so my distance must be incorrect. i tried putting the total distance the pot has fallen, which would be (L_w/t + g*t/2)^2 / ( 2 * g) + h_b and this was incorrect as well. once again i am overlooking something, any help would be appreciated.
     
  11. Sep 7, 2004 #10
    did you sqrt the v_b^2 + 2*g(h_b)? since vf^2 = vi^2 + 2ad
     
  12. Sep 7, 2004 #11

    Doc Al

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    Since the distance (h_b) was given, we must presume it to be correct. Again, it looks like you forgot that your equation gives the final velocity squared. Take the square root to get the velocity. (As needhelpperson points out.)
     
    Last edited: Sep 7, 2004
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