3b. Derivation for maximum height above ground
In this section you will derive an equation that gives the maximum height above the ground your projectile
will reach. You will also include a table of the maximum heights for the three launch angles required.
The formula for the maximum height, $\Delta y_{max}$, should be in terms of:
\begin{itemize}
\item horizontal distance to target, $\Delta x$
\item vertical height to target, $\Delta y$
\item launch angle, $\theta_0$
\item launch height, $h$
\end{itemize}
As with the previous derivation, make sure this one is clearly laid out, showing all steps along the way and any
explanations or assumptions made.
Each cell in Table 4 represents the spring deformation, $\Delta d$, required for the given launch angle and
horizontal distance.
Table 4: Maximum height above ground
Launch Angle, $\theta_0$
Calculations:
30°
45°
60°
Horizontal Distance to Target, $\Delta x$ (m)
2
4
6
8
10
