The solution to a problem is the equation below.Which description best fits this solution?

\frac { 1 } { 2 } \left( 500 \frac { \mathrm { N } } { \mathrm { m } } \right) ( 0.120 \mathrm {~m} ) ^ { 2 } - ( 2.00 \mathrm {~kg} ) \left( 9.80 \frac { \mathrm { m } } { \mathrm { s } ^ { 2 } } \right) ( 0.120 \mathrm {~m} )

= \frac { 1 } { 2 } ( 2.00 \mathrm {~kg} ) \left( 0.824 \frac { \mathrm { m } } { \mathrm { s } } \right) ^ { 2 } + ( 2.00 \mathrm {~kg} ) \left( 9.80 \frac { \mathrm { m } } { \mathrm { s } ^ { 2 } } \right) ( 0.029 0 \mathrm {~m} )

A)A vertical spring compressed 0.120 m shoots a 2.00 kg mass 2.90 cm above the equilibrium position of the spring.
B)A vertical spring stretched 0.120 m shoots a 2.00 kg mass 9.10 cm above the equilibrium position of the spring.
C)A vertical spring compressed 0.120 m shoots a 2.00 kg mass 12.0 cm above the equilibrium position of the spring.
D)A vertical spring compressed 0.120 m shoots a 2.00 kg mass 14.9 cm above the equilibrium position of the spring.
E)A 2.00 kg mass has fallen 0.820 m and compressed the upper end of a vertical spring 12.0 cm below the equilibrium position.

A Skier Weighing 0

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