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

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(D) The maximum height reached by the projectile is $H_{\max} = \frac{u^2 \sin^2 	heta}{2g}$.At half the maximum height,$h = \frac{H_{\max}}{2} = \fr

Vedclass · Accepted Answer

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

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(D) The maximum height reached by the projectile is $H_{\max} = \frac{u^2 \sin^2 	heta}{2g}$.At half the maximum height,$h = \frac{H_{\max}}{2} = \frac{u^2 \sin^2 	heta}{4g}$.Using the equation of motion $v_y^2 = u_y^2 - 2gh$,we find the vertical component of velocity at this height:$v_y^2 = (u \sin 	heta)^2 - 2g \left( \frac{u^2 \sin^2 	heta}{4g} ight) = u^2 \sin^2 	heta - \frac{u^2 \sin^2 	heta}{2} = \frac{u^2 \sin^2 	heta}{2}$.Thus,$v_y = \frac{u \sin 	heta}{\sqrt{2}}$.The power delivered by gravity is $P = \vec{F} \cdot \vec{v} = (mg \hat{j}) \cdot (v_x \hat{i} + v_y \hat{j}) = -mg v_y$ (since gravity acts downwards).So,$P = -\frac{mg u \sin 	heta}{\sqrt{2}}$ (during ascent).During descent,the vertical velocity is downwards,so $P = +\frac{mg u \sin 	heta}{\sqrt{2}}$.Since $\cos(90^\circ + 	heta) = -\sin 	heta$,option $(c)$ is $-\frac{mg u \sin 	heta}{\sqrt{2}}$,which matches the power during ascent. Thus,both $(b)$ and $(c)$ represent the magnitude or specific directional cases of power.

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

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