A point mass $m$ is projected from the earth surface with speed $v_0$ and at an angle $θ$ above the horizontal. Assume that the gravitational acceleration is constant and has the absolute value $g$.
Use energy conservation to show that, at any time, the velocity of the particle is equal in magnitude to the magnitude of the velocity it would acquire in falling freely to that point from a height $\dfrac{v_0^2}{2g}$ above the earth surface.
I have worked out that the kinetic energy $T$ of the projected particle is $$T=\frac{1}{2}m(v_0^2-2v_0gt\sin\theta+g^2t^2)$$
And the potential energy $V$ of the projected particle is $$V(t)=mgtv_0\sin\theta-\frac{1}{2}mg^2t^2$$
Now the initial energy of the falling particle is $$E_0=\frac{v_0^2}{2}m$$
How would I work out the final energy of the particle and then complete the answer to the question?
Thanks in advance.
We note that for our projected particle we have acceleration vector:
$$\vec{a}(t)=\begin{pmatrix}0 \\ -g\end{pmatrix}$$
And thus we have velocity vector:
$$\vec{v}(t)=\begin{pmatrix}v_{0}\cos(\theta) \\ v_{0}\sin(\theta) - gt\end{pmatrix}$$
Integrating again and setting $\vec{s}(0)=\vec{0}$, we get:
$$\vec{s}(t)=\begin{pmatrix}v_{0}\cos(\theta)\:t \\ v_{0}\sin(\theta)\:t-\frac{g}{2}t^{2}\end{pmatrix}$$
We note that:
$$T(t)=\frac{1}{2}m\|\vec{v}(t)\|^{2}=\frac{1}{2}m\left(v_{0}^{2}\cos^{2}(\theta)+v_{0}^{2}\sin^{2}(\theta)-2gtv_{0}\sin(\theta)+g^{2}t^{2}\right) \\ =\frac{1}{2}m\left(v_{0}^{2}+g^{2}t^{2}-2gtv_{0}\sin(\theta)\right)$$
And $$V(t)=mgh=mgs_{y}(t)=mg\left(v_{0}\sin(\theta)t-\frac{g}{2}t^{2}\right)$$
Summing the kinetic and potential energy:
$$E(t)=T(t)+V(t)=\frac{1}{2}m\left(v_{0}^{2}+g^{2}t^{2}-2gtv_{0}\sin(\theta)\right)+mg\left(v_{0}\sin(\theta)t-\frac{g}{2}t^{2}\right) \\=\frac{1}{2}mv_{0}^{2}+\frac{mg^{2}t^{2}}{2}-\frac{mg^{2}t^{2}}{2}=\frac{1}{2}mv_{0}^{2}$$
As required.