Collision Problem: Find Height & Heat Released

[Xsnac]

Homework Statement

a ball of mass 0.1 kg, is free falling from the height (h1) of 1 m. hits the ground with a k = 0.5.
$k=\frac{v_2}{v_1}$
A)Height that the ball will reach after the hit.
B)The heat quantity released by collision

Homework Equations

I think $mgh$ and $\frac{mv^2} {2}$

The Attempt at a Solution

I found $h_2$ = 0,1 m , is this correct? .
divided the process in 2 steps :
$mgh_1= \frac {mv^2_1}{2}$ and found $v_1=\sqrt{2gh_1}$
and then did the same for the other 2 parts of the "movie"
$mgh_2= \frac {mv^2_2}{2}$ and found $h_2=\frac{v_2^2}{2g}$
then I just replaced $v_2$ and calculated $h_2$

 

[mfb]

k=v2v1

I guess that means k=v2/v1.

0.1m is not correct. What did you get as intermediate results?

By the way: decimal point instead of comma in English.

 

[Xsnac]

I guess that means k=v2/v1.

0.1m is not correct. What did you get as intermediate results?

By the way: decimal point instead of comma in English.

well i found $h_2 = \frac {k \sqrt{2gh_1}} {2g}$

 

[mfb]

That equation is wrong (it has wrong units, it cannot work), but if you don't show your individual steps it is impossible to tell what exactly went wrong.

 

[Xsnac]

That equation is wrong (it has wrong units, it cannot work), but if you don't show your individual steps it is impossible to tell what exactly went wrong.

Ok let's see. after the 4 equations I posted on the first thread I found out $h_2$ with the following steps :
$\frac{mv^2_2}{2}=mgh_2$ => $h_2=\frac{v^2_2}{2g}$ , then I replace $v^2_2$ with $(kv_1)^2$ ok I found the error i think ? $h_2=k^2h_1$ ?? should it be 0,25 ?

 

[haruspex]

Ok let's see. after the 4 equations I posted on the first thread I found out $h_2$ with the following steps :
$\frac{mv^2_2}{2}=mgh_2$ => $h_2=\frac{v^2_2}{2g}$ , then I replace $v^2_2$ with $(kv_1)^2$ ok I found the error i think ? $h_2=k^2h_1$ ?? should it be 0,25 ?

Yes.