Throwing Upward

A ball thrown straight up slows down, stops for an instant at the top, then falls back. With up positive, $a = -9.8\ \mathrm{m/s^2}$ the whole time.

The slope is $-9.8\ \mathrm{m/s^2}$ all the way; v = 0 at the top.

At the top $v = 0$, which gives:

$$t_{\text{top}} = \frac{v_0}{g}, \quad h_{\max} = \frac{v_0^2}{2g}$$