Answer
$\textbf{(a)}$ The average speed is
$$v_{ave} =0.98\text{ m/s}$$
$\textbf{(b)}$ The required distance is
$$d=1.52\text{ m}.$$
Work Step by Step
(a) We can calculate the average speed as follows:
$v=\frac{\Delta S}{\Delta t}$
We plug in the known values to obtain:
$v=\frac{2\pi \times 5}{32}$
$v=0.98m/s$
(b) We know that
$t=\sqrt{\frac{-2\Delta y}{g}}$
We plug in the known values to obtain:
$t=\sqrt{\frac{-2(-11.75)}{9.81}}$
$t=1.548s$
Now $d=v_x t$
We plug in the known values to obtain:
$d=(0.982m/s)(1.548s)$
$d=1.52m$
