A particle projected from ground moves at ang | vedclass

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Question

(b)

$	heta=45^{\circ}, t=1 \,s$

$	an \phi=\frac{V_y}{U_y}=\frac{u \sin 	heta-g t}{u \cos 	heta}$

$	an 45^{\circ}=\frac{u \sin 	heta-g \

Vedclass · Accepted Answer

$	an ^{-1}(2)$

Vedclass · Answer

(b)

$	heta=45^{\circ}, t=1 \,s$

$	an \phi=\frac{V_y}{U_y}=\frac{u \sin 	heta-g t}{u \cos 	heta}$

$	an 45^{\circ}=\frac{u \sin 	heta-g 	imes 1}{u \cos 	heta} \Rightarrow u \cos 	heta=u \sin 	heta-g$

also, $V_y=0$, after $1^{	ext {st }}$ (as speed is minimum)

$u \sin 	heta-g 	imes 2=0$

$\Rightarrow u \sin 	heta=2 g \ldots (i)$

so, $u \cos 	heta=2 g-g$

$u \cos 	heta=g \ldots (ii)$

so, $\frac{(i)}{(ii)}=\frac{u \sin 	heta}{u \cos 	heta}=\frac{2 g}{g}$

$\Rightarrow 	an 	heta=2$

$	heta=	an ^{-1}(2)$

A particle projected from ground moves at angle $45^{\circ}$ with horizontal one second after projection and speed is minimum two seconds after the projection. The angle of projection of particle is [Neglect the effect of air resistance]

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