This is a classic example of conservation of energy. Assuming that there are no losses due to friction with air we'll proceed by saying that the total energy mus be conserved.
![E_m=E_k+E_p](https://img.qammunity.org/2019/formulas/physics/college/l41s4d5x7dp0kmn5wucch3qz51luf0uyxc.png)
Now having information on the speed at the lowest point we can say that the energy of the system at this point is purely kinetic:
![E_m=Ek=(1)/(2)mv^2](https://img.qammunity.org/2019/formulas/physics/college/i4eyyh4n7diypm2t1x5hlgjg9a99xmessp.png)
Where m is the mass of the pendulum. Because of conservation of energy, the total energy at maximum height won't change, but at this point the energy will be purely potential energy instead.
![E_m=E_p](https://img.qammunity.org/2019/formulas/physics/college/4wh3l8bgh81kpjpbponkpls7fu52xjjjpz.png)
This is the part where we exploit the Energy's conservation, I'm really insisting on this fact right here but it's very very important, The totam energy Em was
![E_M=(1)/(2)mv^2](https://img.qammunity.org/2019/formulas/physics/college/k0hq783lhqyy9z959bbl7iyoi7by520xhx.png)
It hasn't changed! So inserting this into the equation relating the total energy at the highest point we'll have:
![E_p=mgh=E_m=(1)/(2)mv^2](https://img.qammunity.org/2019/formulas/physics/college/1g2dwpyss93uc94yjbj0pu7s7gp6da17nc.png)
Solving for h gives us:
![h=(v^2)/(2g).](https://img.qammunity.org/2019/formulas/physics/college/drpi04ymhd79hzq08iop7dzm3caicg0xw5.png)
It doesn't depend on mass!