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.

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**Further Reading**

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