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(B) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{u^2 \sin(2	heta)}{g}$,where $u$ is the muzzle speed,$	heta$ is the

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(B) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{u^2 \sin(2	heta)}{g}$,where $u$ is the muzzle speed,$	heta$ is the angle of projection,and $g$ is the acceleration due to gravity.Given: $R_1 = 3.0 \; km$ at $	heta_1 = 30^{\circ}$.Substituting these values: $3.0 = \frac{u^2 \sin(60^{\circ})}{g} = \frac{u^2}{g} \cdot \frac{\sqrt{3}}{2}$.Thus,$\frac{u^2}{g} = \frac{3.0 	imes 2}{\sqrt{3}} = 2\sqrt{3} \approx 3.464 \; km$.The maximum range $R_{	ext{max}}$ for a fixed muzzle speed $u$ occurs at $	heta = 45^{\circ}$,where $R_{	ext{max}} = \frac{u^2}{g}$.Substituting the value of $\frac{u^2}{g}$: $R_{	ext{max}} = 2\sqrt{3} \approx 3.464 \; km$.Since the maximum possible range is approximately $3.46 \; km$,it is impossible to hit a target at $5.0 \; km$ by only adjusting the angle of projection.

$A$ bullet fired at an angle of $30^{\circ}$ with the horizontal hits the ground $3.0 \; km$ away. By adjusting its angle of projection,can one hope to hit a target $5.0 \; km$ away? Assume the muzzle speed to be fixed,and neglect air resistance.

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