#### Answer

$2 \sqrt 5 i+\sqrt 5 j$

#### Work Step by Step

Consider the statement that the velocity vector is tangent to the graph of $y^2=2x$ at the point$(2,2)$ and having length $5$
The direction of the velocity vector is given by
$y^2=2x$ and $2y\dfrac{dy}{dx}=2$
So, we have the slope at point$(2,2)$ will be $\dfrac{dy}{dx}=\dfrac{1}{2}$
Here, we have the tangent vector lies in the direction of the vector , that is, $i+\dfrac{1}{2} y$
Now, velocity $v=5(\dfrac{i+\dfrac{1}{2} y}{\sqrt{1+1/4}})$
Thus $v=2 \sqrt 5 i+\sqrt 5 j$