Understanding kinematics formula in two dimensions Determine the angle of projection of a projectile if its speed at maximum height is $\sqrt{\frac{2}{5}}$ of its speed at half the maximum height.
My solution:
$$H_{max}=\frac{{v_0}^2\sin^2(\theta)}{2g}\implies \frac{1}{2}H_{max}=\frac{{v_0}^2\sin^2(\theta)}{4g}\\v_{x}=v_0\cos(\theta)\quad {v_{\frac{H}{2}y}}^2={v_0}^2\sin^2(\theta)-2g\left(\frac{{v_0}^2\sin^2(\theta)}{4g}\right)=\frac{1}{2}{v_0}^2\sin^2(\theta)\\v_0\cos(\theta)=\sqrt{\frac{2}{5}}\sqrt{{v_0}^2\cos^2(\theta)+\frac{1}{2}{v_0}^2\sin^2(\theta)}\\\cos(\theta)=\sqrt{\frac{2}{5}}\sqrt{\cos^2(\theta)+\frac{1}{2}-\frac{1}{2}\cos^2(\theta)}\\\cos^2(\theta)=\frac{1}{5}\cos^2(\theta)+\frac{1}{5}\\\cos^2(\theta)=\frac{1}{4}\implies\cos(\theta)=\frac{1}{2}\implies\theta=\frac{\pi}{3}$$
Solution found on another website:
$$gH_{max}=\frac{{v_0}^2\sin^2(\theta)}{2}\quad {v_x}^2={v_0}^2\cos^2(\theta)\\{v_{\frac{H}{2}}}^2={v_0}^2-2g\left(\frac{1}{2}H_{max}\right)={v_0}^2-\frac{{v_0}^2\sin^2(\theta)}{2}\\v_{0}\cos(\theta)=\sqrt{\frac{2}{5}}v_{\frac{H}{2}}\implies {v_{0}}^2\cos^2(\theta)=\frac{2}{5}{v_{\frac{H}{2}}}^2=\frac{2}{5}\left({v_0}^2-\frac{{v_0}^2\sin^2(\theta)}{2}\right)\\5\cos^2(\theta)=2-\sin^2(\theta)=1+\cos^2(\theta)\\\cos^2(\theta)=\frac{1}{4}\implies\cos(\theta)=\frac{1}{2}\implies\theta=\frac{\pi}{3}$$
What I don't quite understand in the second solution is the application of the kinematics formula $v^2={v_0}^2+2a\Delta d$ (second line). I thought the formula held only for one dimensional kinematics, but its usage here would imply two dimensional vector addition since the initial velocity and gravity aren't parallel vectors. Can someone help clarify this for me?

You know that
$$v_y^2 = v_{0y}^2 + 2a_y\Delta y$$ and that
$$v_x^2 = v_{0x}^2 + 2a_x\Delta x.$$
First,
$$v^2 = \textbf{v}\cdot\textbf{v} = \left(v_x \hat{\textbf{x}} + v_y \hat{\textbf{y}}\right)\cdot\left(v_x \hat{\textbf{x}} + v_y \hat{\textbf{y}}\right) = v_x^2 + v_y^2.$$
Second,
$$v_0^2 = \textbf{v}_0\cdot\textbf{v}_0 = v_{0x}^2 + v_{0y}^2,$$
and
$$2a_x\Delta x = 0.$$
Last, $$v^2 = v_0^2 - 2g\Delta y.$$
