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(C) For maximum horizontal range,the angle of projection is $	heta = 45^{\circ}$.The horizontal range is given by $R = \frac{u^2 \sin(2	heta)}{g} =

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$\left( \frac{R}{2}, \frac{R}{4} \right)$

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(C) For maximum horizontal range,the angle of projection is $	heta = 45^{\circ}$.The horizontal range is given by $R = \frac{u^2 \sin(2	heta)}{g} = \frac{u^2}{g}$.The speed of a projectile is minimum at the highest point of its trajectory,where the vertical component of velocity becomes zero and only the horizontal component $u_x = u \cos 	heta$ remains.The coordinates of the highest point are $(x, y) = \left( \frac{R}{2}, H ight)$.The maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.Substituting $	heta = 45^{\circ}$,we get $H = \frac{u^2 \sin^2 45^{\circ}}{2g} = \frac{u^2 (1/2)}{2g} = \frac{u^2}{4g}$.Since $R = \frac{u^2}{g}$,we can write $H = \frac{R}{4}$.Therefore,the coordinates of the point where the speed is minimum are $\left( \frac{R}{2}, \frac{R}{4} ight)$.

$A$ projectile is thrown into space so as to have maximum horizontal range $R$. Taking the point of projection as the origin,the coordinates of the point where the speed of the particle is minimum are:

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