Step 1
The ball is thrown into air with a velocity of 40 ft/s. The equation of its position (y) is given by: \[y = 40t - 16{t^2}\]

Step 2: (a)
The average velocity for time period from t=2 and lasting $h$ s is given by ratio of change in position and time taken:

\[\begin{array}{l}{V_{avg}}{\rm{ = }}\dfrac{{{\rm{Change}}\,\,{\rm{in}}\,\,{\rm{Velocity}}}}{{{\rm{Time}}\,\,{\rm{taken}}}}\\{\rm{ = }}\dfrac{{y\left( {2 + h} \right) - y\left( 2 \right)}}{{\rm{h}}}\\{\rm{ = }}\dfrac{{\left[ {40\left( {2 + h} \right) - 16{{\left( {2 + h} \right)}^2}} \right] - \left[ {40\left( 2 \right) - 16{{\left( 2 \right)}^2}} \right]}}{{\rm{h}}}\\{\rm{ = }}\dfrac{{40\left( {2 + h} \right) - 16\left( {4 + 4h + {h^2}} \right) - 80 + 64}}{{\rm{h}}}\\{\rm{ = }}\dfrac{{80 + 40h - 64 - 64h - 16{h^2} - 80 + 64}}{{\rm{h}}}\\{\rm{ = }}\dfrac{{40h - 64h - 16{h^2}}}{{\rm{h}}}\end{array}\]

Step 3: (i)$h=0.5$
\[\begin{array}{l}{{\rm{V}}_{avg,0.5}}{\rm{ = }} - 24 - 16\left( {0.5} \right)\\{\rm{ = }} - 24 - 8\\{\rm{ = }} - 32\end{array}\]Therefore,
\[{{\rm{V}}_{avg,0.5}}{\rm{ = }} - 32\]

Step 4: (ii)$h=0.1$
\[\begin{array}{l}{{\rm{V}}_{avg,0.1}}{\rm{ = }} - 24 - 16\left( {0.1} \right)\\{\rm{ = }} - 24 - 1.6\\{\rm{ = }} - 25.6\end{array}\]Therefore,
\[{{\rm{V}}_{avg,0.1}}{\rm{ = }} - 25.6\]

Step 5: (iii)$h=0.05$
\[\begin{array}{l}{{\rm{V}}_{avg,0.05}}{\rm{ = }} - 24 - 16\left( {0.05} \right)\\{\rm{ = }} - 24 - 0.8\\{\rm{ = }} - 24.8\end{array}\]Therefore,
\[{{\rm{V}}_{avg,0.05}}{\rm{ = }} - 24.8\]

Step 6: (iv)$h=0.01$
\[\begin{array}{l}{{\rm{V}}_{avg,0.01}}{\rm{ = }} - 24 - 16\left( {0.01} \right)\\{\rm{ = }} - 24 - 0.16\\{\rm{ = }} - 24.16\end{array}\]Therefore,
\[{{\rm{V}}_{avg,0.01}}{\rm{ = }} - 24.16\]

Step 7: (b)
As we can observe that, with decrease in the time interval ($h$), the velocity is approaching -24. So Instantaneous velocity is: Therefore,
\[{{\rm{V}}_{Inst}}{\rm{ = }} - 24\,\,{\rm{ft/s}}\]