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(D) Let the ground particle be $P_1$ and the particle from height $h$ be $P_2$.For $P_1$,the time of flight is $T = \frac{2u \sin 	heta}{g}$. The max

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$4H$

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(D) Let the ground particle be $P_1$ and the particle from height $h$ be $P_2$.For $P_1$,the time of flight is $T = \frac{2u \sin 	heta}{g}$. The maximum height reached is $H = \frac{u^2 \sin^2 	heta}{2g}$,which implies $u^2 \sin^2 	heta = 2gH$.For the particles to strike in the air,the time taken by $P_2$ to fall from height $h$ must be less than or equal to the time of flight of $P_1$.For $P_2$,the time taken to fall a vertical distance $h$ is $t = \sqrt{\frac{2h}{g}}$.For the collision to occur,we must have $t \leq T$.Substituting the values: $\sqrt{\frac{2h}{g}} \leq \frac{2u \sin 	heta}{g}$.Squaring both sides: $\frac{2h}{g} \leq \frac{4u^2 \sin^2 	heta}{g^2}$.$h \leq \frac{2u^2 \sin^2 	heta}{g}$.Since $H = \frac{u^2 \sin^2 	heta}{2g}$,we have $u^2 \sin^2 	heta = 2gH$.Substituting this into the inequality: $h \leq \frac{2(2gH)}{g} = 4H$.Thus,the maximum height $h$ is $4H$.

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