Answer Each Question Appropriately $g$ (Length-Units) $\mathrm { sec } ^ { 2 }$

Answer each question appropriately.
-The position of an object in free fall near the surface of the plane where the acceleration due to gravity has a constant magnitude of $g$ (length-units) / $\mathrm { sec } ^ { 2 }$ is given by the equation:
$\mathrm { s } = - \frac { 1 } { 2 } \mathrm { gt } ^ { 2 } + \mathrm { v } _ { 0 } \mathrm { t } + \mathrm { s } _ { 0 }$ , where $\mathrm { s }$ is the height above the earth, $\mathrm { v } _ { 0 }$ is the initial velocity, and $\mathrm { s } _ { 0 }$ is the initial height. Give the initial value problem for this situation. Solve it to check its validity. Remember the positive direction is the upward direction.

A) $\frac { \mathrm { d } ^ { 2 } \mathrm {~s} } { \mathrm { dt } { } ^ { 2 } } = - g , \mathrm {~s} ^ { \prime } ( 0 ) = \mathrm { v } _ { 0 } , \quad \mathrm {~s} ( 0 ) = \mathrm { s } 0$
B) $\frac { \mathrm { d } ^ { 2 } \mathrm {~s} } { d \mathrm { t } ^ { 2 } } = - g \mathrm { t } , \mathrm { s } ( 0 ) = \mathrm { s } 0$
C) $\frac { \mathrm { d } ^ { 2 } \mathrm {~s} } { \mathrm { dt } ^ { 2 } } = \mathrm { g } , \mathrm { s } ^ { \prime } ( 0 ) = \mathrm { v } _ { 0 } , \quad \mathrm {~s} ( 0 ) = \mathrm { s } 0$
D) $\frac { \mathrm { d } ^ { 2 } \mathrm {~s} } { \mathrm { dt } ^ { 2 } } = - \mathrm { g }$