In your answers to this problem, use m for meters, and s for seconds. The formula gives the vertical position of an object, at time $t$, thrown with an initial velocity $v_0$, from an initial position $y_0$ in a place where the acceleration of gravity is $a$. What is the height of a baseball thrown with an initial velocity of $v_0={60\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}$, from an initial position of $y_0= {79\ {\rm m}}$, and at time $t={3\ {\rm s}}$? Note that the acceleration of gravity on earth is ${-9.8\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s^{2}}}}$. It is negative, because we consider the upward direction as positive in this situation, while the gravity pulls the object down.