Maximum Trajectory Area...

The equation of trajectory is :
$\displaystyle y=x\tan{\alpha}\cdot(1-\frac{x}{R})$
where, $\alpha$ is the angle of projection, and, $R$ is the horizontal range.
By, integrating this equation w.r.t. $x$ within the limit $[0,R]$ , we obtain the area of trajectory, $A$ .
$\displaystyle\therefore A=\int^{R}_0y\,\mathrm{d}x=\frac{1}{6}R^2\tan{\alpha}$
But, $R=\frac{v_0^2}{g}(\sin{2\alpha})$ .
So, in order to maximize $A$ , we need to maximize $\displaystyle \sin^2{2\alpha}\cdot\tan\alpha$ .
Now, finding the first derivative of this expression (w.r.t. $\alpha$ ) and equating it with $0$ , we obtain, $\tan^2{\alpha}=3$
$\therefore \alpha=60^{\circ}$ .

I posted similar problem !!

here the horizontal range will be [(V^2* COS* SIN)/g] and the vertical range will be [(V^2* SIN^2)/2/g] now here the motion formed by the thrown object is half ellipse or parabola.Lets assume the motion to be half parabola for simplicity.So area it will occupy is [(PI* v^4* SIN^3* COS)/8/g^2].it is half parabola so area of it is [(PI* half major axis* half minor axis)/2] so the number 8 appeared. Now here except [COS*SIN^3] other components are constants.Now take COS=x and so SIN=(1-x^2)^1/2. Now solve the function for x for maximum value.

integrate the equation of trajectory from 0-R.. (R = Range) then differentiate the obtained expression w.r.t (theta). the differentiation is a little mathematical , you have to intentionally remove the extraneous root sinx=0