A ball dropped from a height of 10m loses 40% of its energy after impact on ground. How much height will it gain after one impact?
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To determine the height the ball will gain after one impact, we need to understand the concept of energy conservation. Energy conservation states that the total energy of a system remains constant, assuming no external forces (such as air resistance or friction) are acting on the system.
In this case, the ball initially has a certain amount of gravitational potential energy due to its height. When it impacts the ground, it loses 40% of its energy, which we can assume is converted into other forms like sound or heat. Therefore, only 60% of the initial energy is available to be converted back into potential energy as the ball rises back up.
To calculate the height gained after one impact, we can use the notion that the potential energy of an object is directly proportional to its height above a reference point (usually chosen as the ground level). The proportionality constant is the gravitational potential energy, which is given by the equation:
PE = m * g * h
where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above the reference point.
Since we are only interested in the change in height, we can compare the initial potential energy (PE_initial) to the final potential energy (PE_final) after the ball rises back up. The ratio of these two energies is the same as the energy retained, which is 60%.
PE_final / PE_initial = 0.6
Since the mass and acceleration due to gravity are constant, they cancel each other out in the equation. Therefore, we can rewrite the equation as:
h_final / h_initial = 0.6
Now we can plug in the values given in the question:
h_initial = 10m (height from which the ball was dropped)
Solving for h_final:
h_final = h_initial * 0.6
= 10m * 0.6
= 6m
Therefore, after one impact, the ball will gain a height of 6 meters.