Answer
a) See the detailed answer below.
b) See the graph below.
Work Step by Step
a)
We know that the ball's initial vertical velocity is 19.6 m/s and that the ball goes under the free-fall acceleration from the moment it leaves the hands of the thrower.
The free-fall acceleration opposes its velocity until it stops and falls back again.
We can find the ball's velocity by using the kinematic formula of velocity
$$v_{y}=v_{0y}+a_yt=19.6 -gt$$
Thus,
$$v_{y}= 19.6 -9.8t\tag 1$$
And we can find the height by using the kinetic formula of position
$$y=y_0+v_{0y}t+\frac{1}{2}a_yt^2$$
We chose the throwing point to be the zero height at which $y_0=0$ m. Thus,
$$y=19.6 t-\frac{1}{2}gt^2\tag 2$$
Using (1) and (2) at the following times.
$\Rightarrow$ At $t=1$ s
$$v_y=\color{red}{\bf 9.8}\;\rm m/s$$ and $$y=\color{red}{\bf14.7}\;\rm m$$
$\Rightarrow$ At $t=2$ s
$$v_y=\color{red}{\bf 0}\;\rm m/s$$ and $$y=\color{red}{\bf
19.6}\;\rm m$$
$\Rightarrow$ At $t=3$ s
$$v_y=\color{red}{\bf -9.8}\;\rm m/s$$ and $$y=\color{red}{\bf
14.7}\;\rm m$$
$\Rightarrow$ At $t=4$ s
$$v_y=\color{red}{\bf -19.6}\;\rm m/s$$ and $$y=\color{red}{\bf
0}\;\rm m$$
b)
Now we need to draw the $v_y-t$ graph from $t=0$ s to $t=4$ s.
