A boy drops a ball from the roof of a house which takes 3 sec to hit the ground. Calculate the velocity before the ball reaches the ground and the height from which it is dropped. (take g =10 m/s/s)

Because the ball is dropped, we know that it starts at a velocity of zero meters per second (fully stopped). 
Gravity is a force of acceleration. As given in the question, we will assume the acceleration force of gravity to be (10 m)/(sec^2) . We can think of this as a constant force of acceleration. Therefore, after one second, the ball is falling at 10 meters per second; at two seconds, it is falling at 20 meters per second; and, at three seconds, it is falling at 30 meters per second. 
The ball is an object in free-fall. It is falling at a constant acceleration, although its velocity and distance are changing. Thus, the instantaneous velocity is simply given by v = g*t
Since the acceleration is uniform, the velocity at the midpoint time t (1.5 seconds) will give us the average velocity. Plugging in 1.5 for the above, we get: 
(10 m)/(sec^2) * 1.5 sec = (15 m)/(sec)
Using this average velocity, we can multiply this with the total time (3 seconds) to obtain 45 meters, the final answer. Thus, the ball was dropped from a height of 45 meters and reached a velocity of 30 meters per second before hitting the ground.
http://www.physicslearningsite.com/uniform-acceleration