This comes from a Saturday Morning Breakfast Cereal (SMBC) comic with a joke answer. The problem states:
A 5 kilogram ball is shot directly right at 20 meters per second from a height of 10 meters. The ball loses 1 joule whenever it touches Earth. Assume no air resistance. When does the ball stop bouncing?
How would one solve this problem? The best I could do was to assume the total energy of the ball, given by the sum of potential and kinetic energy when it's initially shot is completely lost when it stops bouncing. This would give us approximately 1490 bounces, with each bounce slowing the ball down and making it bounce ever so slightly lower.
This still requires a ton of calculation (a huge series), even with the added assumption that there is no friction between the ball and the ground. Am I missing something?
Best Answer
If one assumes that momentum is lost in both directions, since kinetic energy is nondirectional, perhaps it is best to assume that it loses energy in each "direction" proportionally to the sine and cosine of the angle of the bounce. Using this assumption, each bounce causes it to shed 2/3 J from horizontal velocity and 1/3 from vertical potential energy. Under this assumption, there are 1491 bounces, each linearly smaller than the last, coming out finally to roughly 250km in distance.