Derivative function: ball thrown vertically has height s = 2 + 20t - 5t^2

The ball moved after the throw vertically in relation to the ground under the law

s = 2 + 20t - 5t^2 (s = distance from the ground)

a) From what height was the ball thrown?

b) What was the initial speed of the ball movement?

c) Was the ball thrown up or down?

d) How many seconds after the throw the ball was furthest away from the ground?

e) How much was the maximum distance of the ball from the ground?

f) How many seconds after the throw the ball landed on the ground?

g) What was the speed of the ball at the moment of falling?

I did the most basic thing -5*2t+20 = -10t+20 (s')

and from that I get -10 meter/sec^2 (s'') (Acceleration is -10 meter/sec^2 which means that the object was thrown up, is this answer for C and B?)

-10t+20=0 which gives me t=2sec (Ball stops at 2. sec?, answer for F?)

-10*0+20=20 meter/sec (20 meter/sec answer for G?)

But I don't understand how can I answer the other questions, what do I have to do?