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Is There Anyone There Who Could Answer This please

  1. Jul 19, 2007 #1
    Is There Anyone There Who Could Answer This...please!!!

    1. The problem statement, all variables and given/known data
    A ball is thrown downward at 5 m/s from a roof 100 m high. What is the velocity when it reaches the ground?
    A stone is dropped from cliff after it has fallen 30 m. What is its velocity?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 19, 2007 #2
    Vf^2=Vi^2+2ad for both questions

    Vf= final velocity
    Vi= initial velocity
     
  4. Jul 19, 2007 #3

    HallsofIvy

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    There are no formulas or anything? If this is homework, surely you must know SOMETHING about it!

    You must show some work. In fact, one difficulty with your not giving any attempt at all is that we don't know what you have to work with. I can think of several ways of doing this problem, ranging from plugging numbers into a given formula to solving a differential equation, but don't know which of them is appropriate for your class.
     
  5. Jul 19, 2007 #4
    Here is the equation used to try to answer it...

    1. A stone is dropped at a cliff after it has fallen 30 m. What is the velocity?

    Given: Vi=0; d= 30 m; Vf=? g = -9.8 m/s^2; sqrt(Vf )= sqrt(Vi + 2 g d)

    sqrt(Vf) = sqrt(0 + 2 (-9.8 m/s^2) (30 m))
    = sqrt(-588 m)

    =24.25 m/s


    2. A ball is thrown downward at 5 m/s from a roof 100 m high. What is the velocity when it reaches the ground?

    Given: g = -9.8 m/s^2; d = 100 m; Vf = ?; Vi= 0;

    Vf = 5 m/s + (2 (-9.8 m/s^2) (100m))
    = 25 m/s + 1960 m/s
    = 1985 m/s

    3. A stone is dropped from a cliff 490 m above its base. How long does the stone take to fall?

    Given: t = ?; Vi = 0 m/s; d = 490 m; g = 9.8 m/s^2

    d = (Vi t )+ ((g (t^2))/2 )

    490 m= 0 + ((9.8 m/s^2 (t^2))/2)
    490 m = 4.9 m/s^2 (t^2)
    490m/4.9 m/s^2 = t^2
    sqrt(100 m^2/s^2) = sqrt(t^2)

    10 m/s=t
     
  6. Jul 20, 2007 #5

    andrevdh

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    With these problems one need to choose a positive direction (direction of positive y axis?). All kinematic quantities are then assigned either positive or negative values according to their directions.

    You implicitly chose upwards as positive by assigning g = -9.8 m/s^2

    This means that downwards directions (and velocities) will be negative. Like in the first problem d = - 30 meters.

    For these problems it might be easier to choose downwards as positive.
     
  7. Jul 20, 2007 #6

    HallsofIvy

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    Well, of course, the left side is not "sqrt(Vf)", just Vf itself. You are, in effect, using "conservation of energy". Initially the potential energy (relative to the base of the cliff) is -9.8*30 times mass and kinetic energy is 0. At the bottom the potential energy is 0 so all that energy has been converted to kinetic energy.

    Yes, that is correct.


    Vf= Vi+ 2gd ??
    I certainly don't recognize this formula! It doesn't even have consistent "units" (2 (-9.8 m/s^2) (100m)) has units of m^2/s^2, not m/s.
    Better to use the same method you did before, except that now the initial speed is 5 m/s: total energy at the top is (1/2)mVi^2+ mgd while total energy at the bottom is (1/2)mVf^2: (1/2)mVf^2= (1/2)mVi^2+ mgd so
    Vf^2= Vi^2+ 2gd.

    Good - except that (490m/ 4.9 m/s^2 is 100 s^2, not 100 m^2/s^2- meters cancel, not multiply). Of course that means that the answer is
    t= 10 s, not 10 m/s which is a good thing since this is supposed to be a time!
     
  8. Jul 23, 2007 #7
    Isn't that just basic SUVAT.

    i.e.

    v=u+at

    v^2=u^2+2as

    s=ut+(1/2at^2)

    s+vt-(1/2at^2)

    where:
    a = acceleration (9.81)
    u = inital velocity
    v = final velocity
    t = time
    s = displacement

    did you guys uses conservation of energy?...why?
     
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