To find the speed with the given values we can apply the energy conservation equations for which we have that the increase in potential energy is compensated in the decrease of the kinetic energy or vice versa.
Since there is conservation and part of the balance we have to
![KE = PE](https://img.qammunity.org/2020/formulas/physics/high-school/6d35cokjgo5b8akya2oi7khhp64om0qbv8.png)
Where,
KE = Kinetic Energy
PE = Potential Energy
The values for this energy are given as
![(1)/(2)mv^2 = mgh](https://img.qammunity.org/2020/formulas/physics/high-school/dlz24jomhu7oiwhvg025hgfsvx8jzb99zg.png)
![(1)/(2) v^2 = gh](https://img.qammunity.org/2020/formulas/physics/college/wbhhyuof1c8e5zys2pkrpoo3eg35d00j0e.png)
![v = √(2gh)](https://img.qammunity.org/2020/formulas/physics/middle-school/s1pl6sgihodp2uhjy5jv5lpgki75wcufir.png)
![v = √(2(9.8)(0.829))](https://img.qammunity.org/2020/formulas/physics/college/pvfojzdekme9t5u7bu6cp0qc5gj5sgqviz.png)
![v = 4.03m/s](https://img.qammunity.org/2020/formulas/physics/college/u9y5mvvp8b63oehtsp57gf17efbzahq5ri.png)
Therefore the speed at the bottom of the swing is 4.03m/s