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Here are two different methods to solve this kind of problem.

1) Use one of the formulae for constant acceleration. In this case, vf2 = vi2 + 2as, where vf is the final velocity, vi is the initial velocity (zero in this case), a is the acceleration (9.8 meters / second2), and s is the distance.

2) Do an energy calculation, as follows: Calculate the potential energy at a height of 6 meters, with the formula PE = mgh. Since we can assume that the entire potential energy gets converted to kinetic energy just before the ball hits the ground, solve for velocity, in the kinetic energy formula.

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Why velocity is non zero when ball dropped to the ground?

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If it was thrown horizontally or dropped, and hit the ground 3.03 seconds later, then it hit the ground moving at a speed of 29.694 meters (97.42-ft) per second. If it was tossed at any angle not horizontal, and hit the ground 3.03 seconds later, we need to know the direction it was launched, in order to calculate the speed with which it hit the ground.


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The ball dropped from 4m height has more kinetic energy just before it hits the ground because it has a higher velocity due to falling from a greater height. Kinetic energy is directly proportional to both mass and the square of velocity, so the ball dropped from 4m height will have more kinetic energy than the one dropped from 2m height.


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