Question

You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. (a) what fraction of its initial energy is lost during the bounce? (b) what is the ball's speed just as it leaves the ground after the bounce? (c) where did the energy go?

Answer 1

a) At a position of 2.0m, the Initial energy is all made up of the potential energy=m*g*hi
and meanwhile at 1.5 all its energy is also potential energy=m*g*hf 

The percentage of energy remaining is E=m*g*hi/m*g*hf x 100 

and since mass and gravity are constant so it leaves us with just E=hi/hf 
which 1.5/2.0 x100= 75% so we see that we lost 25% of the energy or 0.25 in fraction 

b) Here use the equation vf^2=vi^2+2gd 

where g is gravity, vf is the final velocity and vi is the initial velocity while d is the distance travelled

so in here we are looking for the vi so let us isolate that variable 
we know that at maximum height or peak, the velocity is 0 so vf is 0 

therefore,

vi =sqrt(-2gd) 
vi =sqrt(-2x-9.81x1.5) 
vi =5.4 m/s

c) The energy was converted to heat due to friction with the air and the ground.