A crane, moving at the steady velocity of 5 m/s, is lifting a load of
bricks when one brick falls off 6 m above the ground.  Dercribe the
motion of the free brick by sketching x vs.t with the ground as the
reference point.
a.What is the greatest height the brick reaches? 
b. how long does it take to reach the ground?
c. what is its speed just before in hits the ground?

This looks like homework... but I'll offer some help anyway.

To answer these three questions you need to use the constant
acceleration equations. Setting upward motion as positive, you have:
x = 6m
V1 = 5m/s
a = -9.8m/s^2

a) To find the max height, set V2 = 0 (the brick stops moving for a
split second when it reaches its max height). Then use the formula
V2^2 = V1^2 + 2ax. Solve for x and then add 6m to your answer to find
the max height.

b) To find the time use x = V1t + 0.5at^2 and solve for t.
c) Use V2^2 = V1^2 + 2ax where v1 = 5m/s ,a=-9.8m/s^2 and x=6m

Best of luck
jpariag-ga

Thank you for your help, but i dun get the formula of #a, after i
solved it v2 also =o, so wat cani do next?

could u check the answer 4 me?
a. 7.28m
b.7.75m
c.9.62m/s

Your answer for (a) is correct, but (b) is way too high.
Think about it: acceleration is roughly 10 m/s downwards. The brick
starts off at 5 m/s upwards, so after 1 second it will be going at
about 5 m/s downwards - which means it will also be at about the same
height it started off at, i.e. about 6m above the ground. Now if it
falls for another second its downwards velocity will increase from
about 5 m/s to about 15 m/s, for an average velocity during the second
second of about 10m/s downwards; but it only has another 6m to fall.
So the total time must be less than 2 seconds.

I also get a different answer for (c) - larger than yours. Note that
you can also check your results for (c) by using your answer for (b)
and the formula v2 = v1 + at.