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(B) The equation of the trajectory of a projectile is given by $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Let the point of projection be $(

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$	an ^{-1}\left(\frac{8}{3}ight)$

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(B) The equation of the trajectory of a projectile is given by $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Let the point of projection be $(0, 0)$. The maximum height $H = 20 	ext{ m}$ occurs at $x = 30 	ext{ m}$.The range $R = 30 	ext{ m} + 10 	ext{ m} = 40 	ext{ m}$.Using the formula for maximum height $H = \frac{u^2 \sin^2 	heta}{2g}$ and range $R = \frac{u^2 \sin 2	heta}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.Dividing $H$ by $R$: $\frac{H}{R} = \frac{u^2 \sin^2 	heta / 2g}{2u^2 \sin 	heta \cos 	heta / g} = \frac{	an 	heta}{4}$.Thus,$	an 	heta = \frac{4H}{R}$.Substituting the values $H = 20 	ext{ m}$ and $R = 40 	ext{ m}$:$	an 	heta = \frac{4 	imes 20}{40} = \frac{80}{40} = 2$.Wait,let's re-examine the trajectory equation $y = x 	an 	heta (1 - \frac{x}{R})$.At $x = 30 	ext{ m}$,$y = 20 	ext{ m}$.$20 = 30 	an 	heta (1 - \frac{30}{40}) = 30 	an 	heta (1 - 0.75) = 30 	an 	heta (0.25) = 7.5 	an 	heta$.$	an 	heta = \frac{20}{7.5} = \frac{200}{75} = \frac{8}{3}$.Therefore,$	heta = 	an^{-1}(\frac{8}{3})$.

The angle of projection of a projectile whose path is shown in the given figure is

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