A cricket fielder can throw the cricket ball with a speed $v_o$. 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 if the time of flight is $\frac{2v_0\sin \theta }{g}$ ?

(D) find $\theta$ at which he should throw the ball that would maximise the horizontal range as found $\frac{v_0}{g}[2u\sin \theta +v_0\sin 2\theta ]$

(E) how does for maximum range change if $u>v_o,u=v_o,u<v_o$ ?

(F) As for maximum range change, if $u=v_0,\theta ={60}^{\circ };u>v_0,{\theta }_{max}=\frac{\pi }{2};u=v_0,\theta ={60}^{\circ }$.
How does compare with that for $u=0(i.e{.45}^{\circ })$

(A)
Step: 1 Draw a rough sketch of the given situation.


Step:2 Write the components of velocity in $x$ and $y$ direction.
As the cricket fielder runs with velocity $u$ (in the horizontal direction) and he throws the ball while running, So the horizontal component of the ball includes his personal speed $u$.
Initial velocity in $x-$direction, $u_x=u+v_0\cos \cos \theta$
Initial velocity in $y-$direction, $u_y=v_0\sin \theta$⁡
where $\theta =$angle of projection

Step:3 Find the angle of projection.
Formula Used: $\tan \theta =\frac{u_y}{u_x}$
Now, the angle of projection with horizontal seen by spectator will be
$\tan \theta =\frac{u_y}{u_x}=\frac{v_0\sin \theta }{u+v_0\cos \theta }\Rightarrow \theta ={\tan }^{-1}\left(\frac{v_0\sin \theta }{u+v_0\cos \theta }\right)$
⁡
Final Answer: $\theta ={\tan }^{-1}\left(\frac{v_0\sin \theta }{u+v_0\cos \theta }\right)$⁡

(B)
Step: 1 Draw a rough sketch of the given situation.


Step:2 Find the time of flight of the ball.
Formula Used: $s=ut+\frac{1}{2}a{t}^2$
As net displacement along $y-$axis is zero over time period $T$ (time of flight).
So, $y=0$
Vertical component of velocity, $u_y=v_0\sin \theta$⁡
Acceleration $a_y=-g$
Time $t=T$
From second equation of motion, $y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$\Rightarrow 0=v_0\sin \sin \theta T+\frac{1}{2}(-g)T^2$
$\Rightarrow T[v_0\sin \sin \theta -\frac{g}{2}T]=0\Rightarrow T=0,\frac{2v_0\sin \sin \theta }{g}$
$T=0$, corresponds to point $O$ only not for any other during the motion.
Hence, $T=\frac{2v_0\sin \theta }{g}$

Final Answer: $T=\frac{2v_0\sin \theta }{g}$

(C)
Formula Used: $R=u_{x}T$

Solution:
Step: 1 Draw a rough sketch of the given situation.


Step:2 Find the horizontal range of the ball.
Formula Used: $R=u_{x}T$
Initial velocity in $x-$direction,$u_x=u+v_0\cos \cos \theta$
Initial velocity in $y-$direction, $u_y=v_0\sin \theta$⁡
The time of flight, $T=\frac{2v_0\sin \theta }{g}$
As the horizontal range is given by $R=u_{x}T$
Substituting the values, we get
$R=(u+v_0\cos \theta )T=(u+v_0\cos \theta )\frac{2v_0\sin \theta }{g}$
$R=\frac{v_0}{g}[2u\sin +v_0\sin 2\theta ]$

Final Answer: $\frac{v_0}{g}[2u\sin +v_0\sin 2\theta ]$

(D)
Step: 1 Draw a rough sketch of the given situation.


Step:2 Find the angle at which horizontal range is maximum.
Given horizontal range $R=\frac{v_0}{g}[2u\sin +v_0\sin 2\theta ]$
As we know that horizontal range is occur when $\frac{dR}{d\theta }=0$
$\Rightarrow \frac{v_0}{g}[2u\cos \theta +v_0\cos 2\theta \times 2]=0$
$\because \frac{v_0}{g}\ne 0$
$\therefore [2u\cos \theta +v_0\cos 2\theta \times 2]=0$
$\Rightarrow 2u\cos \theta +2v_0[2{\cos }^2-1]=0$
$\Rightarrow 4v_0{\cos }^2\theta +2u\cos \theta -2v_0=0$
$\Rightarrow 2v_0{\cos }^2\theta +u\cos \theta -v_0=0$
Using quadratic formula, we get
$\Rightarrow \cos \theta =\frac{-u\pm \sqrt{u^2+8v_0^2}}{4v_0}$
$\Rightarrow {\theta }_{max}={\cos }^{-1}\left[\frac{-u\pm \sqrt{u^2+8v_0^2}}{4v_0}\right]$

Final Answer: ${\theta }_{max}={\cos }^{-1}\left[\frac{-u\pm \sqrt{u^2+8v_0^2}}{4v_0}\right]$

(E)
Step: 1 Draw a rough sketch of the given situation.


Step:2 Find the maximum angle at which horizontal range is maximum.
Given horizontal range $R=\frac{v_0}{g}[2u\sin \theta +v_0\sin 2\theta ]$
As we know that horizontal range is occur when $\frac{dR}{d\theta }=0$
$\Rightarrow \frac{v_0}{g}[2u\cos \theta +v_0\cos 2\theta \times 2]=0$
$\because \frac{v_0}{g}\ne 0$
$\therefore [2u\cos \theta +v_0\cos 2\theta \times 2]=0$
$\Rightarrow 2u\cos \theta +2v_0[2{\cos }^2\theta -1]=0$
$\Rightarrow 4v_0{\cos }^2\theta +2u\cos \theta -2v_0=0$
$\Rightarrow 2v_0{\cos }^2\theta +u\cos \theta -v_0=0$
Using quadratic formula, we get
$\Rightarrow \cos \cos \theta =\frac{-u\pm \sqrt{u^2+8v_0^2}}{4v_0}\dots \left(i\right)$

Step: 3 Find the angle when $u=v_0$.
If $u=v_o$, then form equation $\left(i\right)$
$\cos =\frac{-v_0\pm \sqrt{v_0^2+8v_0^2}}{4v_0}$
$\cos \cos \theta =\frac{-1\pm 3}{4}$
$\cos \cos \theta =\frac{1}{2},-1$
Rejecting negative value, as $\theta \ne {180}^{\circ }$
$\Rightarrow \theta =\left(\frac{1}{2}\right)$
$\Rightarrow \theta ={60}^{\circ }$

Step: 4 Find the angle when $u>v_0$.
From equation $\left(i\right)$
$\cos \cos \theta =\frac{-u\pm \sqrt{u^2+8v_0^2}}{4v_0}$
$\cos \theta =\frac{-u\pm u}{4v_0}$
As $u>v_0$
$\therefore \frac{-u\pm u}{4v_0}\to 0$
$\therefore \cos \theta =0=\cos {90}^{\circ }$
${\theta }_{max}=\frac{\pi }{2}$

Step: 5 Find the angle when $u<v_0$.
From equation $\left(i\right)$
$\cos \cos \theta =\frac{-u\pm \sqrt{u^2+8v_0^2}}{4v_0}$
If $u<v_o$ , or $\frac{u}{v_o}<1$
${\theta }_{max}={\cos }^{-1}\left[\frac{\pm 2\sqrt{2}}{4}\right]={\cos }^{-1}[\pm \frac{1}{\sqrt{2}}]$
Rejecting negative value,
${\theta }_{max}={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{4}$

Final Answer: when $u=v_0,\theta ={60}^{\circ }$
when $u>v_0,{\theta }_{max}=\frac{\pi }{2}$
when $u<v_0,{\theta }_{max}=\frac{\pi }{4}$

(F)
Step: 1 Draw a rough sketch of the given situation.


Step:2 Compare $\theta$.
If $u=0,{\theta }_{max}={\cos }^{-1}\left[\frac{0\pm \sqrt{8v_0^2}}{4v_0}\right]$
${\theta }_{max}={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)={45}^{\circ }$
Hence, when $u=0,\theta \le {45}^{\circ }$

Final Answer: $\theta \le {45}^{\circ }$