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(B) The total horizontal range $R$ is the distance from the point of projection to the point where the ball falls. Given the wall is at $x = 6 \, m$ a

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$	an^{-1}\left(\frac{2}{3}ight)$

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(B) The total horizontal range $R$ is the distance from the point of projection to the point where the ball falls. Given the wall is at $x = 6 \, m$ and the ball falls $18 \, m$ beyond the wall,the total range is $R = 6 + 18 = 24 \, m$.The equation of the trajectory is given by $y = x 	an 	heta \left(1 - \frac{x}{R}ight)$.Here,$y = 3 \, m$ at $x = 6 \, m$,and $R = 24 \, m$.Substituting these values into the equation:$3 = 6 	an 	heta \left(1 - \frac{6}{24}ight)$$3 = 6 	an 	heta \left(1 - \frac{1}{4}ight)$$3 = 6 	an 	heta \left(\frac{3}{4}ight)$$3 = \frac{18}{4} 	an 	heta$$3 = 4.5 	an 	heta$$	an 	heta = \frac{3}{4.5} = \frac{30}{45} = \frac{2}{3}$Therefore,$	heta = 	an^{-1}\left(\frac{2}{3}ight)$.

$A$ ball is thrown from the ground to clear a wall $3 \, m$ high at a distance of $6 \, m$ and falls $18 \, m$ away from the wall. The angle of projection of the ball is:

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