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(A) The ball is projected from a height $h = 10 \ m$ and returns to the same vertical level ($10 \ m$ height). This is equivalent to the horizontal ra

Vedclass · Accepted Answer

$8.66$

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(A) The ball is projected from a height $h = 10 \ m$ and returns to the same vertical level ($10 \ m$ height). This is equivalent to the horizontal range of a projectile on a level plane.The formula for the horizontal range $R$ is given by:$R = \frac{u^2 \sin(2	heta)}{g}$Given:Initial velocity $u = 10 \ m/s$Angle of projection $	heta = 30^\circ$Acceleration due to gravity $g = 10 \ m/s^2$Substituting the values:$R = \frac{(10)^2 \sin(2 	imes 30^\circ)}{10}$$R = \frac{100 	imes \sin(60^\circ)}{10}$$R = 10 	imes \frac{\sqrt{3}}{2}$$R = 5\sqrt{3} \ m$Using $\sqrt{3} \approx 1.732$:$R = 5 	imes 1.732 = 8.66 \ m$

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