$A$ cricket fielder can throw a cricket ball with a speed $v_{0}$. If he throws the ball while running with speed $u$ at an angle $\theta$ to the horizontal,find:
$(a)$ The effective angle to the horizontal at which the ball is projected in the air as seen by a spectator.
$(b)$ The time of flight.
$(c)$ The horizontal range from the point of projection at which the ball will land.
$(d)$ The angle $\theta$ at which he should throw the ball to maximize the horizontal range found in $(c)$.
$(e)$ How does $\theta$ for maximum range change if $u > v_{0}$,$u = v_{0}$,and $u < v_{0}$?
$(f)$ How does $\theta$ in $(e)$ compare with that for $u = 0$ (i.e.,$45^{\circ}$)?

(D) Let the velocity of the fielder be $\vec{u} = u\hat{i}$ and the velocity of the ball relative to the fielder be $\vec{v}_{rel} = v_{0}\cos\theta\hat{i} + v_{0}\sin\theta\hat{j}$.
$(a)$ The velocity of the ball relative to the ground is $\vec{v} = (u + v_{0}\cos\theta)\hat{i} + (v_{0}\sin\theta)\hat{j}$. The effective angle $\alpha$ is given by $\tan\alpha = \frac{v_{0}\sin\theta}{u + v_{0}\cos\theta}$,so $\alpha = \tan^{-1}\left(\frac{v_{0}\sin\theta}{u + v_{0}\cos\theta}\right)$.
$(b)$ The vertical component of velocity is $v_{y} = v_{0}\sin\theta$. The time of flight $T$ is given by $T = \frac{2v_{y}}{g} = \frac{2v_{0}\sin\theta}{g}$.
$(c)$ The horizontal range $R$ is $R = v_{x}T = (u + v_{0}\cos\theta)\left(\frac{2v_{0}\sin\theta}{g}\right) = \frac{2uv_{0}\sin\theta + v_{0}^{2}\sin(2\theta)}{g}$.
$(d)$ To maximize $R$,set $\frac{dR}{d\theta} = 0$: $\frac{d}{d\theta}\left[\frac{2uv_{0}\sin\theta + v_{0}^{2}\sin(2\theta)}{g}\right] = 0 \Rightarrow 2uv_{0}\cos\theta + 2v_{0}^{2}\cos(2\theta) = 0 \Rightarrow v_{0}\cos(2\theta) = -u\cos\theta$.
$(e)$ Solving $v_{0}(2\cos^{2}\theta - 1) = -u\cos\theta$,we get $2v_{0}\cos^{2}\theta + u\cos\theta - v_{0} = 0$. Using the quadratic formula for $\cos\theta$: $\cos\theta = \frac{-u + \sqrt{u^{2} + 8v_{0}^{2}}}{4v_{0}}$. As $u$ increases,$\cos\theta$ decreases,meaning $\theta$ increases.
$(f)$ For $u=0$,$\theta = 45^{\circ}$. If $u > 0$,$\theta < 45^{\circ}$ to compensate for the added horizontal velocity $u$.