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(B) Given: Initial speed $u = 50 \ m/s$,angle of projection $	heta = 53^{\circ}$,and height $y = 75 \ m$.Using the trajectory equation of a projectil

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(B) Given: Initial speed $u = 50 \ m/s$,angle of projection $	heta = 53^{\circ}$,and height $y = 75 \ m$.Using the trajectory equation of a projectile: $y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$.Taking $g = 10 \ m/s^2$,$	an 53^{\circ} = \frac{4}{3}$,and $\cos 53^{\circ} = \frac{3}{5}$.Substituting the values: $75 = x \left(\frac{4}{3}ight) - \frac{10 x^2}{2 	imes (50)^2 	imes (3/5)^2}$.$75 = \frac{4x}{3} - \frac{10 x^2}{2 	imes 2500 	imes (9/25)}$.$75 = \frac{4x}{3} - \frac{10 x^2}{1800} 	imes 25 = \frac{4x}{3} - \frac{x^2}{180}$.Multiplying by $180$: $13500 = 240x - x^2$.Rearranging into a quadratic equation: $x^2 - 240x + 13500 = 0$.Solving for $x$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{240 \pm \sqrt{57600 - 4(13500)}}{2} = \frac{240 \pm \sqrt{57600 - 54000}}{2} = \frac{240 \pm \sqrt{3600}}{2} = \frac{240 \pm 60}{2}$.$x_1 = \frac{300}{2} = 150 \ m$ and $x_2 = \frac{180}{2} = 90 \ m$.The horizontal separation is $\Delta x = |x_1 - x_2| = 150 - 90 = 60 \ m$.

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