Let's assume that the initial height is zero. If not, nothing changes but the question b.: what is the maximum height reached over the initial height?
Denote the initial velocity as `V_0.` It is positive and upward. The velocity of the ball as a function of time is
and the height of the ball is
where `g approx 10m/s^2` is the gravity acceleration.
The maximum height is reached when `V(t_1)` is zero. It is going to be negative after that moment, so the ball starts to fall.
at `t_1=V_0/(g)=3(s).` This is the answer to the question a.
b. The maximum height is `H(t_1)=V_0*V_0/(g)-(g*(V_0)^2/(g^2))/2=(V_0^2)/(2g).`
In this case it is 900/20=45 (m).
c. The time `t_2` when the ball returns to the origin is when `H(t_2)=0,` i.e.
The root `t_2=0` corresponds to the initial moment while `t_2=(2V_0)/(g)` =6(s) is the return time. It is no surprise that `t_2=2t_1` because the path of the ball is symmetrical.
d. `V(t_2)=V_0-g*t_2=V_0-g*(2V_0)/(g)=-V_0` =-30(m/s).
This means 30m/s downwards, also no surprise.
Of course all above is true if we ignore the air resistance (which is not entirely correct).