Near the surface of the earth, objects are are subjected to a downward gravitational acceleration denoted by $g$, with value 32 $\mathrm{ft}\/{\mathrm{s}}^2$ (9.8 $\mathrm{m}\/{\mathrm{s}}^2$). It is this gravitational acceleration that accounts for the force pulling objects earthward.

Establish a coordinate system with $y$ positive upward, measured from ground level. Then, the problem is described by the following.

$y″\left(t\right)\=-32$, $y\prime \left(0\right)\=60$, $y\left(0\right)\=500$ 

As in Example 3.10.2, the position function $y\left(t\right)$ is found by antidifferentiating twice, paying attention to the initial conditions $y\prime \left(0\right)\=60$ and $y\left(0\right)\=500$. Hence, $y\prime \left(t\right)\=-32 t\+c$, with $c\=60$ to satisfy the condition $y\prime \left(0\right)\=60$, so $y\left(t\right)\=-32 t\+60$.

As in Example 3.10.2, use the AntiderivativePlot command to obtain the required antiderivatives. After obtaining the position as a function of time, render it a function by selecting the Assign Function option in the Context Panel.

The answers to the remaining three questions can be extracted from the position function $y\left(t\right)$.