Acceleration due to gravity on earth is $\displaystyle {9.8}\frac{{m}}{{s}^{{2}}}={g}\frac{{m}}{{s}^{{2}}}$. The polynomial function for determining the height of a projectile is given by, $\displaystyle {h}{\left({t}\right)}={y}_{{0}}+{v}_{{0}}{t}-\frac{{1}}{{2}}{g}{t}^{{2}}$. Here $\displaystyle {y}_{{0}}$ is the starting height of the projectile and $\displaystyle {v}_{{0}}$ is the initial vertical velocity.

On a tower 312 meters tall a ball is shot upwards with an initial vertical velocity of 112 $\displaystyle \frac{{m}}{{s}}$. What will be its height, relative to the ground floor, after 12 seconds?

You must round your answer to two significant digits because all of the data given to you contains at least two significant digits and acceleration on earth (9.8 meters per second per second) has exactly two significant digits.

The ball will be high after 12 seconds, relative to the ground floor.

Include the unit of measurement in your answer. Include commas when appropriate (for example 18 million is written 18,000,000).