Originally Posted by

**craig** Thanks for the reply.

For the velocity, $\displaystyle \frac{1}{2}mv^2 = 3mg.$, so $\displaystyle v = \sqrt{6g} \approx 7.668$

Just to clarify, $\displaystyle h_f$ is the initial height?

no, $\displaystyle \textcolor{red}{h_f}$ is **f**inal height

So from your equation, $\displaystyle v_f^2 = v_0^2 + 2a(\Delta y)$.

So $\displaystyle 6g + 2(14 - g)(\Delta y) = 0$

$\displaystyle \Delta y = \frac{-6g}{2(14 - g)} = \frac{-3g}{14 - g} = -7$

So the further distance fallen is 7 metres?

correct