Rewrite $\sin \left(\tan ^{-1} \frac{v}{2}\right)$ as an algebraic expression in $v$.
\[
\sin \left(\tan ^{-1} \frac{v}{2}\right)=
\]

Step 1 ：Let's rewrite \(\sin \left(\tan ^{-1} \frac{v}{2}\right)\) as an algebraic expression in v.

Step 2 ：The expression inside the sine function is the inverse tangent of v/2. We can use the identity \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\) to rewrite the expression.

Step 3 ：If we let \(\theta = \tan^{-1}(\frac{v}{2})\), then \(\tan(\theta) = \frac{v}{2}\).

Step 4 ：We can draw a right triangle with opposite side v, adjacent side 2, and hypotenuse \(\sqrt{v^2 + 4}\).

Step 5 ：Then, \(\sin(\theta) = \frac{v}{\sqrt{v^2 + 4}}\).

Step 6 ：\(\boxed{\sin \left(\tan ^{-1} \frac{v}{2}\right)= \frac{v}{\sqrt{v^2 + 4}}}\) is the final answer.