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The 1200 Kg open cage of a tower drop amusement park ride is shown in the picture to the right. The cage has a potential energy of 705,600 J when it is at the top of the drop. Ignoring friction, at what height during the cage drop would the cage have a velocity of 13 m/s? A 51 m B 8.6 m C 60 m D 45 m

Physics
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best answer could be 8.6m
what formula did u use?
use conservation of energy and solve for mgh=0.5mv^2; expression is h= v^2/(2g)

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Other answers:

so u dont need to use the mass at all?
the masses cancel out eventually..
how so?
in all cases when dealing with energy, and asking for the height, or in this one specifically?
mgh=0.5mv^2 it is a general expression for the height when u r using the principle of conservation of energy. potential energy equals change in kinetic energy. therefore we get an equation for the height h= V^2/ 2g
oh so it would most probably always cross out! ahh thank you!
yup.. :) u welcome

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