Answer
$v = \sqrt{g~\lambda}$
Work Step by Step
The wave speed $v$ is measured in units of $m~s^{-1}$
$g$ is measured in units of $m~s^{-2}$
$\lambda$ is measured in units of $m$
Let's assume that $v = g^a~\lambda^b$
Then: $m~s^{-1} = (m~s^{-2})^a~(m)^b$
We can consider the units of $s$:
$(s^{-2})^a = s^{-1}$
$s^{-2a} = s^{-1}$
$-2a = -1$
$a = \frac{1}{2}$
We can consider the units of $m$:
$(m)^a~(m)^b = m^1$
$a+b = 1$
$b = 1-a$
$b = 1-\frac{1}{2}$
$b = \frac{1}{2}$
We can use the exponents to write the equation:
$v = g^a~\lambda^b$
$v = g^{1/2}~\lambda^{1/2}$
$v = \sqrt{g~\lambda}$
