Projectile has just been Maximized !

With $v$ being the initial launch speed we have that $x = v\cos(\theta)t$ and $y = v\sin(\theta)t - \dfrac{gt^{2}}{2}.$

So with $D$ being the distance of the projectile from $P$ we have that

$D^{2} = x^{2} + y^{2} = v^{2}t^{2}\cos^{2}(\theta) + v^{2}t^{2}\sin^{2}(\theta) + \dfrac{g^{2}t^{4}}{4} - vg\sin(\theta)t^{3}$

$\Longrightarrow D^{2} = v^{2}t^{2} - vg\sin(\theta)t^{3} + \dfrac{g^{2}t^{4}}{4}.$

Differentiating with respect to time $t$ we find that

$2D\dfrac{dD}{dt} = 2v^{2}t - 3vg\sin(\theta)t^{2} + g^{2}t^{3} \Longrightarrow \dfrac{dD}{dt} = \dfrac{t}{2D}(g^{2}t^{2} - 3vg\sin(\theta)t + 2v^{2}).$

So $\dfrac{dD}{dt} \gt 0$ as long as $g^{2}t^{2} - 3vg\sin(\theta)t + 2v^{2} \gt 0.$ SInce this quadratic opens upward, we know that it will always be positive as long as its discriminant is negative, leading to the inequality

$(3vg\sin(\theta))^{2} - 4(g^{2})(2v^{2}) \lt 0 \Longrightarrow v^{2}g^{2}(9\sin^{2}(\theta) - 8) \lt 0 \Longrightarrow \sin^{2}(\theta) \lt \dfrac{8}{9} \Longrightarrow \sin(\theta) \lt \dfrac{2\sqrt{2}}{3},$

where we took the positive root as clearly $0^{\circ} \lt \theta \lt 90^{\circ}.$

Since $\sin(\theta)$ increases from $0^{\circ}$ to $90^{\circ},$ in order for $\dfrac{dD}{dt}$ to always be positive we will thus require that

$\theta \lt \sin^{-1}\left(\dfrac{2\sqrt{2}}{3}\right) = \boxed{70.53^{\circ}}$ to 2 decimal places.

Hint : distance will increases till ,

\displaystyle{\vec { V } \^ \vec { r } \le { 90 }^{ 0 }\\ \quad \left\{ at\quad max: \right\} \\ \Rightarrow \boxed { \vec { V } .\vec { r } =0 } }

Answer is :

$\displaystyle{\theta =\sin ^{ -1 }{ \left( \sqrt { \cfrac { 8 }{ 9 } } \right) } ={ (70.53) }^{ 0 }}$