How long is it in the air?
Realize that time to go up = time to go down.
At the top of the flight, Velocity=zero
Vf=Vo+gt
0 = 20m/s + gt
-20m/s = -9.8m/s/s(t)
t = 2.04 seconds to go up
So total time in the air is 4.1 seconds
Note: Often you will see "g" expressed as a negative number.  This accounts for gravity acting downward.  So if you shoot something upward, you are moving it in the "+" direction.

What is its maximum height?
We know it returns to the surface at 20m/s.
Let's choose to consider the down part of the trip (we could have just as easily chosen up).
It starts from rest, so Vo=zero, leaving us with...
Vf² = 2gx
x = Vf²/2g = (400m²/s²)/19.6m/s² = 20.4m

How high is it at time = 1 second?

DX = Vot + at²/2 = (20m/s*1s) + ((-9.8m/s/s*1s²)/2) = 15.1m

Points to ponder...
* the 15.1m answer to the last question is the same answer to this question: "How far will any object be above the ground after one second if dropped from h=20m?"
*Doesn't this seem weird: If an object is dropped from 20m up, it is only in the air for ~2seconds AND that after one of those seconds it has only fallen about 5meters?  Strange but true.
 

What if the object in the above problems were thrown at an angle of 60° with the ground?

Now, instead of all of the strength of the throw going into the upward motion of the ball, some of it goes into sideways motion of the ball.

The initial velocity stays the same, 20m/s.
What changes?
Since the ball is going at an angle, the upward velocity is less than 20m/s.
What is it (upward Velocity) then?
We are looking for the upward component. Imagine a triangle...the angle between the ball's path and the Y axis is 30°, so the Y component can be found using cos30=y/20, or
y = 20cos30 = 17.32m/s which is the same as 20sin60 = 17.32m/s

What is its sideways (often called "X") velocity?

Looking at the triangles again, now we want the side adjacent to the 60° angle, so...
X = 20cos60 = 10m/s  which is the same as 20sin30 = 10m/s

Another point to ponder...
Think for a second; the throw angle was 60 degrees.  That is a pretty high angle.  So it should make sense that the X speed came out lower than the Y speed.  In other words, more oomph went into the up than into the across.
Can you guess the angle that would give you the farthest throw?

Where would this ball be after 1 second?

We must do 2 problems.
(1) Find the height, using the Y velocity
(2) Find the horizontal distance using the X velocity

#1: Y = 17.32m/s*1sec + (-9.8m/s/s*(1s)²)/2 = 12.42m up

#2: X= vt = 10m/s*1s = 10m horizontally (remember X motion is constant; no gravity to worry about)

*When a projectile is at the top of its trajectory, it may seem to "sit there" for a second before it starts to move back down.  This is especially apparent when a projectile is shot straight up... it seems to hover for a time before it starts to come back down.
It is true that the projectile may have an instantaneous speed of zero, BUT the object is still in the air, so it is still affected by gravity and STILL HAS an acceleration of 9.8m/s/s.  All projectiles are accelerated during their whole flight!