Question 724419
If that projectile is fired from ground level, its height (in meters) t seconds after being shot would be give by the equation
{{{h=588t-4.9t^2}}} (the 4.9 number is half of the acceleration of gravity in {{{m/s^2}}} ).
Solving {{{-4.9t^2+588t=0}}} would give us the time when it hits the ground.
Dividing both sides of the equation by {{{-4.9}}} we get
{{{t^2-120t=0}}} so the projectile is at ground level at t=0 (we knew that) and at t=120, {{{highlight(120 seconds)}}}.