Question 1119884
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A ball is dropped from a height of 20.0 ft. above the ground
(a) At what height will one half of its energy be kinetic and one half potential?
(b) Using only energy considerations, what is the velocity of the ball just as it strikes the ground?
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(a) At what height will one half of its energy be kinetic and one half potential?


<pre>
    The initial potential energy was P = m*g*h,   

    where m is the mass, g = 32 ft/s^2 is the gravity acceleration and h = 20 ft.


    When the falling body reaches the height of 10 meters above the ground, its potential energy will be half

    of the original potential energy.


    The other half of the initial potential energy will transform to the kinetic energy of the falling body.


    Hence, at the the height of 10 meters above the ground, the body will have equal amounts of its potential and kinetic energy.


    It gives the <U>Answer</U> to the problem question:


        At the height of 10 meters above the ground, the body will have equal amounts of potential and kinetic energy.

        Its total energy remains the same, but now the half of the energy is potential, while the another half is kinetic.
</pre>


(b) Using only energy considerations, what is the velocity of the ball just as it strikes the ground?


<pre>
    {{{(m*V^2)/2}}} = {{{mgh}}}  ====>  {{{V^2}}} = 2*g*h  ====>  V = {{{sqrt(2*g*h)}}} = {{{sqrt(2*32*20)}}} = {{{sqrt(1280)}}} = 35.78 ft/s.
</pre>

Solved.