(a) As the cricket fielder runs with velocity $u$ (in horizontal direction) and he throws the ball, while running horizontal component of ball includes his initial speed $u$ in $X$-direction is given by
So, $u_x=u_0+v_0(\cos \theta)$
vertical component of velocity of ball does not affect.
Le, $u_y=$ initial velocity in $y$-direction
$u_y=v_0 \sin \theta$
where angle of projection is $\theta$.
Now, we can write
$$
\begin{aligned}
&\tan \theta=\frac{u_y}{u_x}=\frac{u_0 \sin \theta}{u+v_0 \cos \theta} \\
&\theta=\tan ^{-1}\left(\frac{v_0 \sin \theta}{u+v_0 \cos \theta}\right)
\end{aligned}
$$
(b) Let us consider $T$ be the time of flight motion As net displacement is zero over time period $T$. $s_y=0$ (as ball return in same positionof y-axis) $u_y=u_0 \sin \theta, a_y=-g, t=T$
We know that, $s_y=u_y t+\frac{1}{2} a_y t^2$
So, $0=v_0 \sin \theta T+\frac{1}{2}(-g) T^2$ $T\left[v_0 \sin \theta-\frac{g}{2} T\right]=0 \Rightarrow T=0, \frac{2 v_0 \sin \theta}{g}$ $T=0$, corresponds to point $O$.
Hence, $T=\frac{2 v_0 \sin \theta}{g}$
(c) Horizontal range,
$$
\begin{aligned}
R &=(u \times T)=\left(u+v_0 \cos \theta\right) T \\
&=\left(u+v_0 \cos \theta\right)\left[\frac{2 v_0 \sin \theta}{g}\right] \\
R &=\frac{v_0}{g}\left[2 u \sin \theta+v_0 \sin 2 \theta\right]
\end{aligned}
$$
(d) For horizontal maximum range, $\frac{d R}{d \theta}=0$
$$
\begin{aligned}
&d\left[\frac{v_0}{g} \frac{2 u \sin \theta+v_0 \sin 2 \theta}{d \theta}\right]=0 \\
&\frac{v_0}{g}\left[2 u \cos \theta+v_0 \cos 2 \theta \times 2\right]=0 \\
&\frac{v_0}{g} \neq 0 \\
&\therefore 2 u \cos \theta+2 v_0\left[2 \cos ^2 \theta-1\right]=0 \\
&4 v_0 \cos ^2 \theta+2 u \cos \theta-2 v_0=0 \\
&2 v_0 \cos ^2 \theta+u \cos \theta-v_0=0
\end{aligned}
$$
By quadratic formula, where $a=2 v_0, b=u, c=\left(-v_0\right)$
$$
\begin{aligned}
&\cos \theta=\frac{-u \pm \sqrt{u^2+8 v_0^2}}{4 v_0} \\
&\theta_{\max }=\cos ^{-1}\left[\frac{-u \pm \sqrt{u^2+8 v_0^2}}{4 v_0}\right] \\
&\text { or } \theta=\cos ^{-1}\left[\frac{-u \pm \sqrt{u^2+8 v_0^2}}{4 v_0}\right]
\end{aligned}
$$
(e)
$$
\begin{aligned}
&\cos \theta=\frac{-u \pm \sqrt{u^2+81 b^2}}{4 v_0} \\
&\cos \theta=\frac{-v_0 \pm \sqrt{v_0^2+8 v_0^2}}{4 v_0}=\left[\frac{-1 \pm 3}{4}\right]
\end{aligned}
$$
As $\theta$ acute angel $\theta$ is angle of projection,
$$
\begin{aligned}
&\cos \theta=\frac{-1+3}{4}=\frac{1}{2} \\
&\cos \theta=\cos 60^{\circ} \\
&\text { so, } \theta=60^{\circ}
\end{aligned}
$$
If $u< < v_0$, then $8 v_0^2+u^2=8 v_0^2$
$$
\theta_{\max }=\cos ^{-1}\left[\frac{-u \pm 2 \sqrt{2} v_0}{4 v_0}\right]=\cos ^{-1}\left[\frac{1}{\sqrt{2}}-\frac{u}{4 v_0}\right]
$$
If $u \ll v_0$, then $\theta_{\max }=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}$
If $u \ll u_0$ and $u \gg v_0$
$$
\begin{aligned}
&\theta_{\max }=\cos ^{-1}\left[\frac{-u \pm u}{4 v_0}\right]=0 \\
&\theta_{\max }=\frac{\pi}{2}
\end{aligned}
$$
(f) If $u=0, \quad \cos \theta=\left[\frac{-4 \pm \sqrt{u^2+8 v_0^2}}{4 v_0}\right]$, so
$$
\theta_{\max }=\cos ^{-1}\left[\frac{0 \pm \sqrt{8 v_0^2}}{4 v_0}\right]=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ}
$$
$$
\theta=\frac{\pi}{4}
$$