I would appreciate it if someone could help me with the following problem. A ball is thrown at an angle of 45 degrees at a speed of 40m/s. Each time it bounces, it loses 75% of its speed. How far does it travel before it stops? What I don't understand is that because it only loses 75% of its speed, doesn't that mean it never stops? If someone could explain a solution to me, I'd be grateful.

The each time it bounces loses 75% of its speed is a data, depend on the floor material and the ball material. We decompose the speed in 2 speeds, one forward (->) and one up (/|). Each of these speeds is sqrt (2) * 20 (do the math). Now the ball accelerates downwards (|/) at 10m/s*s. So it takes sqrt (2) * 20/10 = sqrt (2) * 2 seconds to stop going up and start going down. In that time it moved sqrt (2) * 2 * sqrt (2) * 20 = 80m. (sqrt (2) * 2 = time, sqrt (2) * 20 = speed). Since it has constant acceleration it describes a parabola. We know the vertex is between both roots, so that means it hits the ground again at 160m. Now the next time, the speed will be 10m/s and the starting angle will be the same. This means it moves 160m * 1/4, or 40m. We can keep on like this forever since it will never stop, but we do realize that it moves always slower. So it moves 160m * (1+1/4+1/4/4+1/4/4/4.................) = 160 * (1+1/4+1/16+1/64........) Now some math. 1 + 1/x + 1/x2 + 1/x3 + 1/x4.... 1/xn + 1/x n+1) = 1 + 1/x-1. (this applies for all x > 1) So 160m * (1+1/4+1/4/4+1/4/4/4......................) = 160 * (1+1/4+1/16+1/64..........) = 160 * (1+1/3) = 160m * 4/3 = 640m/3. The ball moves 640m/3. Though it actually never stops.

True, that is the closest answer, IMO, the question itself is wrong since the ball would literally never stop on a closed system (and we didn't deal with friction either)

Reminds me of one of my old physics problems, although I don't remember exactly how it was formulated.

Fuck, shamanics answered it again. Well he is 200% right. My all labour solving on my note-book went in vain.