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2 Interesting Questions:

  1. Sep 24, 2006 #1
    I will start with the one that I have worked out already. I am submitting the answers to an online testing site and the answer that I have submitted is wrong but I believe that my math and selection of proper equations is correct.


    A pellet gun is fired straight downward from the edge of a cliff that is 11.4 m above the ground. The pellet strikes the ground with a speed of 27.5 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward?

    MY WORK:

    First we have to find the initial velocity (which will be the same in both situations).

    I used: V^2 = Vo^2 + 2(a)(y)

    Which is 27.5^2 = Vo^2 + 2(-9.80)(11.4)

    Vo = 31.3 m/s

    Then use the equation with V = 0 as it is at the peak of the shot.

    y = V^2 - Vo^2/(2(-9.80))

    -31.30^2/-19.6 = 49.984 m <----------- WRONG ANSWER, anything you see wrong?


    You live in a building on the left, and a friend lives in a building on the right. The two of you are having a discussion about the heights of the buildings, and your friend claims that his building is half again as tall as yours. To resolve the issue you climb to the roof of your building and estimate that your line of sight to the top edge of the other building makes an angle of 21° above the horizontal, while your line of sight to the base of the other building makes an angle of 52° below the horizontal.


    (a) Determine the ratio of the height of the taller building to the height of the shorter building.

    MY WORK:

    I have no Idea where to begin to find the solution to this, I have tried 21/52, that does not work. Any suggestions?

    Thanks so much guys(and girls)
  2. jcsd
  3. Sep 24, 2006 #2

    Andrew Mason

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    For the first question, your method is correct so I think your problem is with the number of significant digits (3). You have 5.

    For the second question, express the height of the shorter house in terms of the separation distance, x and the lower angle. Then express the additional height of the taller in terms of x and the upper angle.

  4. Sep 25, 2006 #3
    if you have the answer, could u pm me? if the answer tallies with mine, i'll post the solution, or reply u in the pm.
  5. Sep 25, 2006 #4

    Andrew Mason

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    Why not see if the OP can get the solution. What trigonometric function would you use here to compare the heights?

  6. Sep 25, 2006 #5


    for the first question, I actually figured it out. For some reason in this case, Acceleration did not need to be negative. hmm. But for the second question, I am still at a loss.

    Can you explain what you mean anymore?

    would it be along the lines of

    H= x(sin 21) + x sin 52

  7. Sep 25, 2006 #6

    Andrew Mason

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    Try H1 = tan(52)/x where x is the horizontal separation. What is H2 (height of taller house)?

  8. Sep 25, 2006 #7

    thanks for all the help Andrew, I wasn't able to get it. the correct answer was 1.30.

    Thanks again though.
  9. Sep 25, 2006 #8
    1st question, the pellet is going downward, its accelerating, so the acceleration wil be positive but not negative
  10. Sep 25, 2006 #9
    Second question
    Let the horizontal line to be H, The opposite line of the angle of the upper part of the triangle to be Y, opposite line of the angle of the lower part of the triangle to be X, the height of the higher building to be X+Y
    use the method of tangent to build up all the equation

    hope i helped, the solution wil get u the final answer
    Last edited: Sep 25, 2006
  11. Sep 25, 2006 #10

    Andrew Mason

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    [tex]H1 = tan(52)/x[/tex]

    [tex]H2 = tan(21)/x + tan(52)/x[/tex]

    [tex]\frac{H2}{H1} = ____?[/tex]

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