$(a)$ Time of ascent is equal to the time of descent. The ball takes a total of $6\, s$ for its upward and downward journey.

Hence, it has taken $3\, s$ to attain the maximum height.

Final velocity of the ball at the maximum height, $v = 0\, m/s$

Acceleration due to gravity, $g = -9.8\, ms^{-2}$

Using equation of motion, $v = u + at$, we have

$0 = u + (-9.8 \times 3)$

$ u = 9.8 \times 3 = 29.4\, m/s$

Hence, the ball was thrown upwards with a velocity of $29.4\, m/s$.

$(b)$ Let the maximum height attained by the ball be $h$.

Initial velocity during the upward journey, $u = 29.4 \,m/s$

Final velocity,$ v = 0 \,m/s$

Acceleration due to gravity, $g = -9.8 \,ms^{-2}$

Using the equation of motion,

$s=u t+\frac{1}{2} a t^{2}$

$h=29.4 \times 3-\frac{1}{2} \times 9.8 \times 3^{2} \quad \Rightarrow h=44.1 \,m$

Hence, the maximum height is $44.1 \,m.$

$(c)$ Ball attains the maximum height after $3\, s$. After attaining this height, it will start falling downwards.

In this case,

Initial velocity, $u = 0\,m/s$

Position of the ball after $4 \,s$ of the throw is given by the distance travelled by it during its downward journey in $4 s – 3 s = 1 s$.

Using the equation of motion, $s=u t+\frac{1}{2} a t^{2}$

$s=0 \times 1+\frac{1}{2} \times 9.8 \times 1^{2} \Rightarrow s=4.9 \,m$

Now, total height $= 44.1 \,m$

This means, the ball is $39.2 \,m\, (44.1 \,m – 4.9\, m)$ above the ground after $4$ seconds.