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5.Work, Energy, Power and Collision
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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
A
$\frac{m g u \cos \theta}{\sqrt{2}}$
B
$\frac{m g u \sin \theta}{\sqrt{2}}$
C
$\frac{m g u \cos (90+\theta)}{\sqrt{2}}$
D
Both $(b)$ and $(c)$
Solution
(d)
$H_{\max }=\frac{u^2 \sin ^2 \theta}{2 g}$
$v_y^2=(u \sin \theta)^2-\frac{2 g u^2 \sin ^2 \theta}{4 g}$
$v_y=\frac{u \sin \theta}{\sqrt{2}}$
At point $A$,
$\vec{P}=\vec{F} \cdot \vec{V}=(m g)\left(\frac{u \sin \theta}{\sqrt{2}}\right) \cos \pi=\frac{-m g u \sin \theta}{\sqrt{2}}$
At point $B$,
$\vec{P}=\frac{+u m g \sin \theta}{\sqrt{2}}$
Standard 11
Physics
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