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(B) For a projectile,the maximum horizontal range $R_{\max}$ is given by $R_{\max} = \frac{u^2}{g}$.Given $R_{\max} = 400 \, m$.The velocity of a proj

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$(200, 100)$

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(B) For a projectile,the maximum horizontal range $R_{\max}$ is given by $R_{\max} = \frac{u^2}{g}$.Given $R_{\max} = 400 \, m$.The velocity of a projectile is minimum at its highest point.At the highest point,the horizontal distance covered is $x = \frac{R}{2} = \frac{400}{2} = 200 \, m$.The maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$. For maximum range,$	heta = 45^\circ$,so $H = \frac{u^2 \sin^2 45^\circ}{2g} = \frac{u^2 (1/2)}{2g} = \frac{u^2}{4g} = \frac{R_{\max}}{4}$.$H = \frac{400}{4} = 100 \, m$.Thus,the coordinates of the highest point are $(x, H) = (200, 100) \, m$.

$A$ projectile is thrown into space so as to have a maximum possible horizontal range of $400 \, m$. Taking the point of projection as the origin,the coordinates of the point where the velocity of the projectile is minimum are

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