A cricket fielder can throw the 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 air as seen by a spectator

$(b)$ what will be time of flight?

$(c)$ what is the distance (horizontal range) from the point of projection at which the ball will land ?

$(d)$ find $\theta$ at which he should throw the ball that would maximise the horizontal range as  found in $(iii)$.

$(e)$ how does $\theta $ for maximum range change if $u > u_0$. $u =u_0$ $u < v_0$ ?

$(f)$ how does $\theta $ in $(v)$ compare with that for $u=0$ $($ i.e., $45^{o})$ ?

Consider the adjacent diagram,

$(a)$ Initial velocity in $x$-direction,

$\mathrm{U}_{x} =\mathrm{U}+\mathrm{V}_{0} \cos \theta$

$\mathrm{U}_{y} =\text { Initial velocity in } y \text {-direction }$

$=\mathrm{V}_{0} \sin \theta$

Where angle of projection is $\theta$.

Now, we can write

$\tan \theta=\frac{U_{y}}{U_{x}}=\frac{U_{0} \sin \theta}{U+U_{0} \cos \theta}$

$\Rightarrow \quad \theta=\tan ^{-1}\left(\frac{V_{0} \sin \theta}{U+V_{0} \cos \theta}\right)$

$(b)$ Let $\mathrm{T}$ be the time of flight,

As net displacement is zero over time period T. $y=0, U_{y}=V_{0} \sin \theta, a_{y}=-g, t=\mathrm{T}$

We know that $y=\mathrm{U}_{y} t+\frac{1}{2} a_{y} t^{2}$

$\Rightarrow \quad 0=\mathrm{V}_{0} \sin \theta \mathrm{T}+\frac{1}{2}(-g) \mathrm{T}^{2}$

$\Rightarrow \quad \mathrm{T}\left[\mathrm{V}_{0} \sin \theta-\frac{g}{2} \mathrm{~T}\right]=0 $

$\Rightarrow \mathrm{T}=0 \frac{2 \mathrm{~V}_{0} \sin \theta}{g}$

$\mathrm{~T}=0_{1} \text {, corresponds to point } \mathrm{O} \text {. }$

Hence, $\mathrm{T}=\frac{2 \mathrm{U}_{0} \sin \theta}{g}$