a 25-g ball is released from rest 80 m above teh surface of the earth. during the fall the total thermal energy of the ball and air increases by 15 j. just before it hits the surface its speed is.

The speed of the ball just before it hits the surface is 20m/s.Here we have to find the speed of the ball just before it hit the surface.So,Mass of the ball = 25g = 0.025kgHeight from which it falls (h) = 80mThermal energy (E) = 15JNow initial energy of the system is [tex]E_{i}[/tex] = mv²/r + mghThe initial velocity is zeroTherefore [tex]E_{i}[/tex] = mgh = 20JNow the final energy of the system is:[tex]E_{f}[/tex] = mv²/r + mgh + 15h = 0 as we have to find the speed when it hit the surface.[tex]E_{f}[/tex] = mv²/2 + 15     = 0.025 v² /2 + 15As there is no external force appliedSo the total energy of the system will be constant,By conservation of energy:[tex]E_{i}[/tex] = [tex]E_{f}[/tex]20 = 0.025 v²/2 + 150.025v² = 10v = 100/5v = 20 m/sTherefore the speed of the ball when it hit the surface is 20m/s.To know more about thermal energy refer to the link given below:https://brainly.com/question/19666326#SPJ4