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rjkz [21]
3 years ago
7

A 0.50 kilogram ball is held at a height of 20 meters. What is the kinetic energy of the ball when it reaches halfway after bein

g released?
Physics
1 answer:
Svetach [21]3 years ago
5 0

While the ball is held 20 meters above the floor,

                                   Potential energy = (mass) (gravity) x (height) =

                                                        (0.5 kg) x (9.8 m/s²) x (20 m) =  98 joules.

When it's dropped and reaches 1/2 of its initial height, it has half as much
potential energy.  The rest has turned into kinetic energy  =  <em>49 joules.

</em>
============================================================<em>
</em>

Want to know what else you can figure out ?  Watch this . . . 

                                                 <u> Kinetic energy = (1/2) x (mass) x (speed)²</u>

                                                          49 joules = (1/2) x (0.5 kg) x (speed)²

Divide each side by (0.25 kg):       49 joules / 0.25 kg = speed²

Take the square root of each side:        7/0.5 = speed = <em><u>14 meters per second</u></em>

That's the speed of the ball after falling 10 meters ! 
How cool is that !

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We must take in consideration that two parts with same mass do not move in same direction. The center of mass of these two parts moves between them at angle of 45° with respect to both south and west. The speed of a center of mass is:
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v_{3f} = - \frac{m*33.9+m*33.9 }{ 2m } \\  \\ v_{3f} = - \frac{2m*33.9 }{ 2m }  \\  \\ v_{3f} = - 33.9m/s

We can see that part of a coconut with biggest mass has same speed as center of mass of two other parts. Negative sign shows that direction is opposite to direction of two pats. Biggest part moves towards north-east.
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