The vertical velocity diver`s,
[tex]v_{y}= \sqrt{2g h}[/tex]
Given [tex]h= 3.40 \ m[/tex]
Therefore,
[tex]v_{y} = \sqrt{2\times 9.8 m/s^2 \times 3.40 \times m } =8.16\ m/s[/tex]
The initial speed is in horizontal direction which is given ,[tex]v_{x} = 1.75 \ m/s[/tex]
We have [tex]v_{x}[/tex] and [tex]v_{y}[/tex], and because this is a right vector triangle, we can use the Pythagorean theorem we can calculate the final velocity.
[tex]v=\sqrt{(v_{x}^2)+(v_{y}^2) } =\sqrt{(1.75)^2+(8.16)^2} = \sqrt{3.0625+66.585}\\\\ v= 8.34 m/s[/tex]
