A ball is thrown vertically downwards at a speed of 10ms^-1 from a height of 10m. Upon hitting the floor 10% of the energy is dissipated through waste heat. What is the heighest point the ball reaches before it comes to rest? Take g=10ms^-2

The first step is to calculate the initial energy of the system. at the moment of throwing. Energy is comprised of both potential (given by Mgh) and kinetic(given by 0.5Mv^2), and so the initial energy is : Mg(10) +1/2 * M * (10)^2 = 150M, where M is mass of the ball and h is its height.As energy is conserved, the energy at the instant of bouncing is the same, so we multiply by 0.9 to get the energy at the start of the next cycle. Again, due to conservation of energy, the highest point is given by the instant when this energy is all converted into potential energy, giving: MgH = 0.9 * 150M , which gives H = 13.5m. Important to note here is that this is the highest point in any of the subsequent cycles, as the total energy in the system can never exceed this value, and as 13.5m > 10m, the answer is 13.5m