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Homework Help: Kinetic energy stuff

  1. Aug 30, 2005 #1
    kinetic energy stuff....

    here's the question...
    an object is thrown upwards at a projection angle of 60 degree with kinetic energy E. When the object reaches max height, its kinetic energy becomes
    A. (1/8)E
    B. (1/4)E
    C. (1/2)E
    D. (3/2)E

    I'm totally no idea....some1 kindly tells me the concept of this question.... x_x
  2. jcsd
  3. Aug 30, 2005 #2
    When the object is just launched it has E Cos (60) energy in the horizontal direction and E Sine (60) in the vertical direction. When it arrives at the top end of travel (vertically) it only has the horizontal energy left.
    Therefore it is E Cos (60) = 0.5 E
  4. Aug 30, 2005 #3

    Doc Al

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    Staff: Mentor

    You need to know what percentage of the object's initial KE is associated with the horizontal component of its velocity. (The part associated with the vertical component will go to zero at max height.) Hint: If the initial speed is V, what's the horizontal component of the velocity?
  5. Aug 31, 2005 #4
    i've seen the answer is 0.25E from the book, so I just think should we squaring the cos60, ie (cos60)^2=0.25?

    Doc Al: if the horizontal component of velocity remains unchange in a projectile motion, so the velocity is still v isn't it?
  6. Aug 31, 2005 #5
    Initial Kinetic energy : E = [itex]\frac{1}{2}mv^2[/itex]

    K.E at the highest position: [itex]\frac{1}{2}m(v cos60)^2[/itex]

    Rest i leave upto you.

  7. Aug 31, 2005 #6
    Brilliant question, makes you see _why_ KE _is_ proportional to velocity squared. If you think about KE has to be, then, this question reveals that it must be proportional to velocity squared simply because it is separable in the different spacial directions. WOW it really blows my mind. It is probably neater to say that the velocity in the horizontal direction is half the initial straight line velocity from simple geometry rather than having to appeal to trig (though trig is more general). And then say it this half velocity that you are squaring to get the quarter energy.
  8. Aug 31, 2005 #7

    Doc Al

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    Staff: Mentor

    Yes, but why? The answer is to compare the initial KE (which comes from both the vertical and horizontal components of the initial velocity) to the final KE (which comes only from the horizontal speed). (See Dr.Brain's post.)

    I'm not sure what you mean. If the initial velocity (with magnitude V) has components:
    [itex]V_{ix} = V \cos \theta[/itex]
    [itex]V_{iy} = V \sin \theta[/itex]

    Then at the top of the motion, since the vertical component is zero, the velocity will have components:
    [itex]V_{fx} = V \cos \theta[/itex]
    [itex]V_{fy} = 0[/itex]
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