Suppose that you throw a ball at an angle to the horizontal, and just after it leaves your hand at a height $y_i$its velocity is$\left(v_{xi},v_{yi},0\right)$. Assuming that we can neglect air resistance, at the top of its trajectory, when it is momentarily traveling horizontally, its velocity is$\left(v_{xi},0,0\right)$. What is the height$y_f$at the top of the trajectory, in terms of the other known quantities? Use the Energy Principle.

There are two instants available.

When the ball is at its highest point, the initial one $y_i$has a speed $V_i=(V_{xi},V_{yi},0)$. The second, when the ball is at its highest point$y_f$ and has a speed$v_f=\left(v_{xi},0,0\right)$

According to the work-energy principle, an increase in the kinetic energy of a rigid body is caused by an equal amount of positive work done on the body by the force acting on it.

There are two instants available. When the ball is at its highest point, the initial one$y_i$and has a speed$V_i=(V_{xi},V_{yi},0)$and the second, when the ball is at its highest point$y_f$and has a speed$V_f=\left\{V_{xi},0,0\right\}$.

The kinetic and potential energies will change in this case, but their sum will remain constant in both instants.

The change in gravitational energy caused by moving the ball is approximately equal to m g y, and we will adopt this approximation because the change in the length is small compared to the earth's radius.

As a result, the principle energy in the form is the same for initial and final states.

To determine the final long, put the expressions of

$$\begin{gathered} K_f+U_f=K_i+U_i\frac{1}{2}m{v}_f^2+mg{y}_f \\ {K}_f+U_f=\frac{1}{2}m{v}_i^2+mg{y}_{i}g{y}_f \\ {K}_f+U_f=\frac{1}{2}\left[v_i^2-v_f^2\right]+g{y}_i{y}_f \\ {K}_f+U_f=\frac{1}{2g}\left[v_i^2-v_f^2\right]+y_i \\ {y}_i=\frac{1}{2g}\left[v_i^2-v_f^2\right]+y_i................\left(1\right) \end{gathered}$$

now let us plug the expressions of $v_{i_{}}$and $v_f$into equation (1) to find the final long $y_f$

$$\begin{gathered} y_f=\frac{1}{2g}\left[v_i^2-v_f^2\right]+y_i \\ {y}_f=\frac{1}{2g}\left[v_{xi}^2+v_{yi}^2-v_x^2\right]+y_i \\ {y}_f=\frac{v_{yi}^2}{2g}+y_i \end{gathered}$$

The change in potential energy is unaffected by movement along the horizontal path, as indicated.

Hence, the height at top of the trajectory$y_f=\frac{v_{yi}^2}{2g}+y_i$.