A sliding conductor reloaded

Let $v$ be the velocity at any point of the movement of the rod. We see that the induced emf is $\epsilon=BLv$ . Then, the current is $I=\dfrac{\epsilon}{R}=\dfrac{BLv}{R}$ and the magnetic force, which goes downwards, has a magnitude of $F_m=ILB=\dfrac{B^2L^2v}{R}$ .

Let $a$ be the acceleration at any point, which will have a direction upwards. So, we have $-F_m-mg=ma$ :

$-B^2L^2v-mgR=mRa$

We know that $a=\dfrac{\mathrm dv}{\mathrm dt}$ and $v=\dfrac{\mathrm dh}{\mathrm dt}$ , hence $a=\dfrac{v\;\mathrm dv}{\mathrm dh}$ and:

$-(B^2L^2v+mgR)=mRv\dfrac{\mathrm dv}{\mathrm dh}$

Separate the variables:

$-mR\dfrac{v\;\mathrm dv}{B^2L^2v+mgR}=dh$

The greatest height will be attained when $v=0$ , so we know the limits of integration.

Use the identity $\dfrac{ax}{bx+c}=\dfrac{a}{b}\left(1-\dfrac{c}{bx+c}\right)$ , so we can integrate that:

$\displaystyle -\dfrac{mR}{B^2L^2}\int_{v_0}^{0} \left(1-\dfrac{mgR}{B^2L^2v+mgR}\right)\mathrm dv=\int_{0}^{h_{max}}\mathrm dh$

$-\dfrac{mR}{B^2L^2}\left[0-v_0-\dfrac{mgR}{B^2L^2}\ln\left(\dfrac{B^2L^2(0)+mgR}{B^2L^2v_0+mgR}\right)\right]=h_{max}$

$h_{max}=\boxed{\dfrac{mR}{B^2L^2}\left[v_0-\dfrac{mgR}{B^2L^2}\ln\left(\dfrac{B^2L^2v_0+mgR}{mgR}\right)\right]}$

Facepalm !! $:-($ Because I didnt think of using $a = v \dfrac{dv}{dh}$ instead just found the time and put it into the height expression. Had to integrate TWICE once for velocity then again for distance. Wasted 2 pages of my book.