A body of mass m is projected from ground wit | vedclass

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Question

(d)

$H_{\max }=\frac{u^2 \sin ^2 	heta}{2 g}$

$v_y^2=(u \sin 	heta)^2-\frac{2 g u^2 \sin ^2 	heta}{4 g}$

$v_y=\frac{u \sin 	heta}{\sqrt{2

Vedclass · Accepted Answer

Both $(b)$ and $(c)$

Vedclass · Answer

(d)

$H_{\max }=\frac{u^2 \sin ^2 	heta}{2 g}$

$v_y^2=(u \sin 	heta)^2-\frac{2 g u^2 \sin ^2 	heta}{4 g}$

$v_y=\frac{u \sin 	heta}{\sqrt{2}}$

At point $A$,

$\vec{P}=\vec{F} \cdot \vec{V}=(m g)\left(\frac{u \sin 	heta}{\sqrt{2}}ight) \cos \pi=\frac{-m g u \sin 	heta}{\sqrt{2}}$

At point $B$,

$\vec{P}=\frac{+u m g \sin 	heta}{\sqrt{2}}$

A body of mass $m$ is projected from ground with speed $u$ at an angle $\theta$ with horizontal. The power delivered by gravity to it at half of maximum height from ground is

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