R त्रिज्या का एक पहिया कीचड़ में फँस गया है औ | vedclass

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Question

$(b)$ Let a mud blob is detached from the circumference of wheel with initial speed $u$ at angle $	heta$ as shown below.

Height upto which mud blo

Vedclass · Accepted Answer

$\frac{u^{2}}{2 g}+\frac{g R^{2}}{2 u^{2}}$

Vedclass · Answer

$(b)$ Let a mud blob is detached from the circumference of wheel with initial speed $u$ at angle $	heta$ as shown below.

Height upto which mud blob can be thrown is

$h=$ Maximum height of projectile $+$ Height at which mud blob is thrown

$	herefore \quad h=\frac{u^{2} \sin ^{2} 	heta}{2 g}+R \cos 	heta \quad \dots(i)$

$h$ is maximum when $\frac{d h}{d 	heta}=0$

$\Rightarrow \frac{d}{d 	heta}\left(\frac{u^{2} \sin ^{2} 	heta}{2 g}+R \cos 	hetaight)$

$\Rightarrow \frac{u^{2}}{2 g} \cdot 2 \sin 	heta \cos 	heta-R \sin 	heta=0$

$\Rightarrow \quad \frac{u^{2}}{g} \cdot \cos 	heta=R 	ext { or } \cos 	heta=\frac{R g}{u^{2}}$

$\Rightarrow \quad \sin ^{2} 	heta=1-\frac{R^{2} g^{2}}{u^{2}}$

Substituting these values in Eq $(i)$, we get

$\lambda_{\max } =\frac{u^{2}}{2 g}\left(1-\frac{R^{2} g^{2}}{u^{4}}ight)+R\left(\frac{R g}{u^{2}}ight)$

$=\frac{u^{2}}{2 g}-\frac{R^{2} g}{2 u^{2}}+\frac{R^{2} g}{u^{2}}=\frac{u^{2}}{2 g}+\frac{R^{2} g}{2 u^{2}}$

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