Hello!

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

`V(t)=V_0-g*t,`

and the height of the ball is

`H(t)=V_0t-(g*t^2)/2,`

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.

`V(t_1)=V_0-g*t_1=0`

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.

`V_0t_2=(g*t_2^2)/2.`

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).

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