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A cricket fielder can throw the cricket ball with a speed vo. 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 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 c)

e) how does $\theta$ for maximum range change if $u>v_{o}=v_{o}$ and $u<v_{o}$

f) how does $\theta$ in e) compare with that for u = 0?

Answers (1)

a) u is the horizontal velocity with which the cricketer runs. The ball is thrown by him while running and hence the speed of ball also contains a component of the cricketer’s speed.

$U_{x}=u+v\; \cos \; \theta$

Vertical component,

$U_{y}=v\; \sin \; \theta$

$\tan \theta =\frac{v\; \sin \theta }{u+v\cos \theta }$

$\theta =\tan ^{-1}[\frac{v\; \sin\; \theta }{u+v\; \cos \; \theta }]$

b) Time of flight

$S_{y}=U_{y}t+\frac{1}{2}a_{y}t^{2}$

Since the ball returns back to the same position, Sy = 0

$U_{y}=v\; \sin \theta$

So, $0=v\; \sin \theta (T)-\frac{1}{2}g\; T^{2}$

$T[v\; \sin \theta -\frac{1}{2}g\; T]=0$

Since T cannot be zero, we have

$T=2v\; \sin \frac{\theta }{g}$

c) Maximum range

for the max range, the condition is

$\frac{dR}{d\theta} =0$

$d\frac{\left \{ \frac{v}{g}[2u]\sin \; \theta +\frac{v}{g}\; \sin 2\theta \right \}}{d\; \theta }=0$

$\theta =\cos ^{-1}\left [ \frac{-u\pm \sqrt{u^{2}+8\; v_{0}^{2}}}{4v_{0}} \right ]$

$\cos \; \theta =\left [ \frac{-u\pm \sqrt{u^{2}+8\; v_{0}^{2}}}{4v_{0}} \right ]$

e) In the case when $u=v,\cos \theta =$

$\frac{-v_{0}\pm \sqrt{v_{0}^{2}+8v_{0}^{2}}}{4v_{0}}$

$=\frac{-v_{0}+3v_{0}}{4v_{0}}$

$\cos\; \theta =-1+\frac{(-3)}{4}$

$\cos\; \theta =\frac{1}{2}$ (as $\theta$ is taken as an acute angle here)

hence, $\theta =60^{o}$

for the case of u << v

$\cos \theta =-u+\frac{(-2\sqrt{2}v)}{4v}$

Since $\theta$ is an acute angle here,

as u << v here, we can neglect the last term

$\cos \theta =\frac{1}{\sqrt{2}}$

$\theta =\frac{\pi }{4}$

For u >> v,

$\cos\; \theta =\frac{-u+(-u)}{4v}$

$\cos\; \theta =0=\cos 90$

$\theta =\frac{\pi }{2}$

f) when $u=0$

$\cos\; \theta =\frac{-u\pm \sqrt{u^{2}+8v_{0}^{2}}}{4v_{0}}$

$\cos\; \theta =\frac{2\sqrt{2}v}{4v}=\frac{1}{\sqrt{2}}$

$\cos\; \theta =45$

$\cos\; \theta =\frac{\pi }{4}$

Posted by

infoexpert23

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