_{Step 1: Find a relation between angles.z}

_{We know that maximum height reached by a projectile,}

_{$\begin{array}{l}\mathrm{H}=\frac{{\mathrm{u}}^2{\sin }^2\mathrm{\theta }}{2\mathrm{g}}\\ {\mathrm{H}}_1=\frac{{\mathrm{v}}_0^2{\sin }^2{\mathrm{\theta }}_1}{2\mathrm{g}} \left(\mathrm{for} \mathrm{first} \mathrm{particle}\right)\\ {\mathrm{H}}_2=\frac{{\mathrm{v}}_0^2{\sin }^2{\mathrm{\theta }}_2}{2\mathrm{g}} \left(\mathrm{for} \mathrm{second} \mathrm{particle}\right)\end{array}$}

_{According to the question, we know that}

_{$\begin{array}{l}\Rightarrow \frac{{\mathrm{H}}_1>{\mathrm{H}}_2}{2\mathrm{g}}>\frac{{\mathrm{v}}_0^2{\sin }^2{\mathrm{\theta }}_1}{2\mathrm{g}}>\frac{{\mathrm{v}}_0^2{\sin }^2{\mathrm{\theta }}_2}{2\mathrm{g}}\\ \Rightarrow {\sin }^2{\mathrm{\theta }}_1>{\sin }^2{\mathrm{\theta }}_2\end{array}$
$\begin{array}{lr}\Rightarrow & {\sin }^2{\mathrm{\theta }}_1-{\sin }^2{\mathrm{\theta }}_2>0\\ \Rightarrow & (\sin {\mathrm{\theta }}_1-\sin {\mathrm{\theta }}_2)(\sin {\mathrm{\theta }}_1+\sin {\mathrm{\theta }}_2)>0\end{array}$
$\begin{array}{lrl}\text{ Thus, either }& \sin {\mathrm{\theta }}_1+\sin {\mathrm{\theta }}_2& >0\\ \Rightarrow & \sin {\mathrm{\theta }}_1-\sin {\mathrm{\theta }}_2& >0\\ \Rightarrow & \sin {\mathrm{\theta }}_1& >\sin {\mathrm{\theta }}_2\text{ or }{\mathrm{\theta }}_1>{\mathrm{\theta }}_2\\ \text{ Time of fight, }& \mathrm{T}& =\frac{2\mathrm{usin}\mathrm{\theta }}{\mathrm{g}}=\frac{2{\mathrm{v}}_0\sin \mathrm{\theta }}{\mathrm{g}}\\ \text{ Thus, }& {\mathrm{T}}_1=\frac{2{\mathrm{v}}_0\sin {\mathrm{\theta }}_1}{\mathrm{g}}& \\ {\mathrm{T}}_2& =\frac{2{\mathrm{v}}_0\sin {\mathrm{\theta }}_2}{\mathrm{g}}& \end{array}$}

_{(Here. ${\mathrm{T}}_1$ = Time of flight of first particle and ${\mathrm{T}}_2$ = Time of flight of second particle).}

Step 2: Find relation between time of flight and the range.

$\begin{array}{lc}\text{ As, }& \sin {\theta }_1>\sin {\theta }_2\\ \text{ Hence, }& T_1>T_2\end{array}$

We know that,

$\begin{array}{l}\text{ Range, }R=\frac{u^2\sin 2\theta }{g}=\frac{v_0^2\sin 2\theta }{g}\\ \begin{array}{c}{A}_1=\text{ Range of first particle }=\frac{u_0^2\sin 2{\theta }_1}{g}\\ R_2=\text{ Range of second particle }=\frac{v_0^2\sin 2{\theta }_2}{g}\end{array}\end{array}$

Given,

$\begin{array}{lc}& \cdot \sin {\theta }_1>\sin {\theta }_2\\ \Rightarrow & \sin 2{\theta }_1>\sin 2{\theta }_2\\ \Rightarrow & \frac{R_1}{R_2}=\frac{\sin 2{\theta }_1}{\sin 2{\theta }_2}>1\\ \Rightarrow & R_1>R_2\end{array}$

Step 3: Find a relation between the total energies.

Total energy for the first particle.

$\begin{array}{l}{\mathrm{U}}_1=\mathrm{KE}+\mathrm{PE}=\frac{1}{2}{\mathrm{m}}_1{\mathrm{v}}_0^2\\ \text{ (This value will be constant throughout the journey) }\\ {\mathrm{U}}_2=\mathrm{KE}+\mathrm{PE}=\frac{1}{2}{\mathrm{m}}_2{\mathrm{v}}_0^2\text{ (Total energy for the second particle) }\end{array}$

Total energy for the particle

$\begin{array}{l}{\mathrm{m}}_1={\mathrm{m}}_2\text{ then }{\mathrm{U}}_1={\mathrm{U}}_2\\ {\mathrm{m}}_1>{\mathrm{m}}_2\text{ then }{\mathrm{U}}_1>{\mathrm{U}}_2\\ {\mathrm{m}}_1<{\mathrm{m}}_2,\text{ then }{\mathrm{U}}_1<{\mathrm{U}}_2\end{array}$