Area made by projectile

If we drop a vertical from the highest point of the projectile take it's intersection with the horizontal origin.

Then equation of curve is of the form $y=b-a{ x }^{ 2 }$

Area of the curve(which is above x-axis) with the x-axis is given by :

$\frac { 2b }{ 3 } \sqrt { \frac { b }{ a } }$

It is easy to note that in our case $b=\frac { { v }^{ 2 } }{ 2g } { { sin }^{ 2 } }\theta$ and $a=\frac { g }{ 2{ v }^{ 2 }{ cos }^{ 2 }\theta }$ (By the equation of trajectory of projectile).

Putting in the values we get $A=\quad \frac { 2{ v }^{ 4 }{ sin }^{ 3 }\theta cos\theta }{ 3{ g }^{ 2 } }$

Calculating $\frac { dA }{ d\theta }$ and equating it to 0 we get $\theta =60\quad degrees$

putting $\theta ={ 60 }^{ 0 }$ we get :

$A=\quad \frac { \sqrt { 3 } }{ 8 } \frac { { v }^{ 4 } }{ { g }^{ 2 } }$

So $a=60,b=3,c=8$ hence $\boxed{a+b+c=71}$

$A\quad =\quad \int _{ 0 }^{ x_{ max } }{ ydx } \\ =\quad \int _{ 0 }^{ 2vsin(\theta )/g }{ \left( (vsin(\theta )t\quad -\quad gt²/2 \right) vcos(\theta )dt } \quad \\ =\quad \frac { 2{ v }^{ 4 } }{ { 3g }^{ 2 } } \sin ^{ 3 }{ \theta \cos { \theta } } \\ The\quad maximum\quad occurs\quad when\quad \tan { \theta } =\sqrt { 3 } ,\quad that\quad is,\quad when\quad \theta =60°.$

Time to reach highest point is say t. Then $t=\dfrac{VSin(a)}{g} \\ H=\dfrac{VSin(a)}{2} *t=\dfrac{VSin(a)}{2} * \dfrac{VSin(a)}{g}\\\therefore~H=\dfrac{V^2Sin^2(a)}{2g}.......................(1)\\Range~R=2* VCos(a )*t=2*\dfrac{V^2Cos(a)*Sin(a)}{g}\\ \therefore~R=2*\dfrac{V^2Cos(a)Sin(a)}{g}.....(2).\\A=\dfrac{2}{3}*H*R.\\From~ (1)~ and~(2)...A=\dfrac{2}{3}*\dfrac{V^4Cos(a)Sin^3(a)}{g^2}\\Differentiating ~ Cos(a)Sin^3(a)~ and ~ equating~ to~ zero,~~\\we~get~tan^2(a)=3\\\implies ~a=60.....A=\dfrac{2}{3}*\dfrac{V^4*\dfrac{1}{2}*(\dfrac{\sqrt3}{2})^3}{g^2} \\a=60, ~~b=3,~~c=8.~~~a+b+c= \boxed{71}$