The equation given is $KE = \frac{1}{2}mv^2$, and you're asked to solve for $v$.

Step-by-step solution:

1. Start with the given equation: $KE = \frac{1}{2}mv^2$

2. Multiply both sides by 2 to eliminate the fraction: $2 \cdot KE = mv^2$

3. Divide both sides by $m$ to isolate $v^2$: $\frac{2 \cdot KE}{m} = v^2$

4. Finally, take the square root of both sides to solve for $v$: $v = \sqrt{\frac{2 \cdot KE}{m}}$

Correct Answer: B

$v = \sqrt{\frac{2KE}{m}}$

Let me know if you'd like further details or clarification.

Relative questions:

1. What happens to $v$ if the kinetic energy $KE$ is doubled?
2. How is the velocity $v$ related to mass $m$ in this equation?
3. What is the physical meaning of kinetic energy in this equation?
4. Can you derive the same equation starting with the formula for velocity and rearranging for $KE$?
5. How would the equation change if it involved potential energy instead of kinetic energy?

Tip: Always remember to double-check your algebraic manipulations when isolating variables, especially when dealing with squares and square roots.