Maximum Area

Let $\theta$ be the angle of projection.

Now, we take the equation of trajectory: $y=x\tan { \left( \theta \right) } -\frac { g{ x }^{ 2 }\sec ^{ 2 }{ \left( \theta \right) } }{ { 2v }^{ 2 } }$

The Range of this motion is: $R=\frac { 2v\sin { \left( \theta \right) } }{ g }$

Now, Area of the parabolic path is: $\displaystyle A=\int _{ 0 }^{ R }{ ydx }$

On, substituting the value of $y$ , we get: $\displaystyle A=\int _{ 0 }^{ \frac { 2v\sin { \left( \theta \right) } }{ g } }{ \left( x\tan { \left( \theta \right) } -\frac { g{ x }^{ 2 }\sec ^{ 2 }{ \left( \theta \right) } }{ { 2v }^{ 2 } } \right) dx }$

This is just integration of linear functions. Therefore area $A=\frac { { 2v }^{ 4 }\sin ^{ 3 }{ \left( \theta \right) } \cos { \left( \theta \right) } }{ 3{ g }^{ 2 } }$

Therefore for maximum area $\boxed{\theta ={ 60 }^{ 0 } }$

Its more of calculus than kinematics!