Tough Energy problem Law of conservation of energy

[Senjai]

Homework Statement

Bill throws a 10.0g (0.0100kg) straight down froma height of 2.0m. The ball strikes the floor at a speed of 7.5 m/s.
a) what was the original speed of the ball?
b) if 30% of the balls energy is transformed in thermal energy during the collision with the floor, find the new height reached by the ball.

Homework Equations

$$\frac{1}{2}mv^2 + mgh = \frac{1}{2}mv'^2 + mgh'$$
$$\sum{E} = \sum{E'}$$

The Attempt at a Solution

A) i solved this fairly easily and got the right answer.

$$v = \sqrt{\frac{2\left(\frac{1}{2}mv'^2 - mgh\right)}{m}}$$

v initial = 4.13 m/s

B) this is the toughy...[/color]

The answer is supposed to be 2.0 m. i attempted to rearrange the question including the initial kinetic energy + initial potential = end potential and solve for h. but i didnt get it. A friend of mine told me i don't need to include work done by NC forces in this equation. i think i got 1.7m, i know that if i just have initial potential energy = to kinetic energy @ the point of collision.. when losing 30% of energy, it isn't supposed to reach the same height correct? Which leads be to believe that the person throwing the ball supplied the additional 30% energy to return it to the 2.0m height. I don't understand how to mathmatically show how to solve to get h' = 2.0m..

Thanks,
Senjai

 

[Delphi51]

I get 2.01 m for (b). I just did
E = .7*.5*m*v^2 = .7*.5*.01*7.5^2 = .1969 J
.28125 = mgh