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(B) Given that the maximum height $H$ is $\frac{1}{4}$ of the horizontal range $R$, so $H = \frac{R}{4}$.Using the formulas $H = \frac{u^2 \sin^2 	he

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$\frac{p^2}{4\,m}$

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(B) Given that the maximum height $H$ is $\frac{1}{4}$ of the horizontal range $R$, so $H = \frac{R}{4}$.Using the formulas $H = \frac{u^2 \sin^2 	heta}{2g}$ and $R = \frac{u^2 \sin(2	heta)}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$, we have:$\frac{u^2 \sin^2 	heta}{2g} = \frac{1}{4} \left( \frac{2u^2 \sin 	heta \cos 	heta}{g} ight)$$\frac{u^2 \sin^2 	heta}{2g} = \frac{u^2 \sin 	heta \cos 	heta}{2g}$$\sin 	heta = \cos 	heta \implies 	an 	heta = 1 \implies 	heta = 45^{\circ}$.The initial momentum is $p = mu$. The horizontal component of velocity is $u_x = u \cos 	heta = u \cos 45^{\circ} = \frac{u}{\sqrt{2}}$.At the highest point, the vertical component of velocity is zero, so the velocity is only the horizontal component $v = u_x = \frac{u}{\sqrt{2}}$.The momentum at the highest point is $p' = m v = m \left( \frac{u}{\sqrt{2}} ight) = \frac{p}{\sqrt{2}}$.The minimum kinetic energy is at the highest point:$K_{	ext{min}} = \frac{(p')^2}{2m} = \frac{(p / \sqrt{2})^2}{2m} = \frac{p^2 / 2}{2m} = \frac{p^2}{4m}$.

An object of mass $m$ is projected with a momentum $p$ at such an angle that its maximum height is $\frac{1}{4}$th of its horizontal range. Its minimum kinetic energy in its path will be

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