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(C) Given: Maximum height $H = 3 \,m$,Range $R = 4 \,m$,and $g = 10 \,ms^{-2}$.We know that $H = \frac{u^2 \sin^2 	heta}{2g}$ and $R = \frac{u^2 \sin

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$10 \sqrt{\frac{2}{3}} \,ms^{-1}$

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(C) Given: Maximum height $H = 3 \,m$,Range $R = 4 \,m$,and $g = 10 \,ms^{-2}$.We know that $H = \frac{u^2 \sin^2 	heta}{2g}$ and $R = \frac{u^2 \sin 2	heta}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.Taking the ratio of $H$ and $R$:$\frac{H}{R} = \frac{u^2 \sin^2 	heta / 2g}{2u^2 \sin 	heta \cos 	heta / g} = \frac{\sin 	heta}{4 \cos 	heta} = \frac{1}{4} 	an 	heta$.Substituting the values: $\frac{3}{4} = \frac{1}{4} 	an 	heta \Rightarrow 	an 	heta = 3$.Since $	an 	heta = 3$,we have $\sin 	heta = \frac{3}{\sqrt{10}}$ and $\cos 	heta = \frac{1}{\sqrt{10}}$.Using the formula for $H$:$3 = \frac{u^2 \sin^2 	heta}{2g} = \frac{u^2 (3/\sqrt{10})^2}{2 	imes 10} = \frac{u^2 	imes (9/10)}{20}$.$3 = \frac{9u^2}{200} \Rightarrow u^2 = \frac{3 	imes 200}{9} = \frac{200}{3}$.$u = \sqrt{\frac{200}{3}} = 10 \sqrt{\frac{2}{3}} \,ms^{-1}$.Thus,the correct option is $C$.

If the maximum height and range of a projectile are $3 \,m$ and $4 \,m$ respectively,then the velocity of the projectile is (Take $g=10 \,ms^{-2}$)

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