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(A) The horizontal range $R$ is given by $R = \frac{u^2 \sin 2	heta}{g}$.Given $R = 5\sqrt{3}\,m$,$u = 10\,m/s$,and taking $g = 10\,m/s^2$:$5\sqrt{3}

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$(\sqrt{3}-1)\,s$

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(A) The horizontal range $R$ is given by $R = \frac{u^2 \sin 2	heta}{g}$.Given $R = 5\sqrt{3}\,m$,$u = 10\,m/s$,and taking $g = 10\,m/s^2$:$5\sqrt{3} = \frac{10^2 \sin 2	heta}{10} \implies 5\sqrt{3} = 10 \sin 2	heta \implies \sin 2	heta = \frac{\sqrt{3}}{2}$.Thus,$2	heta = 60^{\circ}$ or $2	heta = 120^{\circ}$,which gives $	heta_1 = 30^{\circ}$ and $	heta_2 = 60^{\circ}$.The time of flight is given by $T = \frac{2u \sin 	heta}{g}$.For $	heta_1 = 30^{\circ}$,$T_1 = \frac{2 	imes 10 	imes \sin 30^{\circ}}{10} = 2 	imes 0.5 = 1\,s$.For $	heta_2 = 60^{\circ}$,$T_2 = \frac{2 	imes 10 	imes \sin 60^{\circ}}{10} = 2 	imes \frac{\sqrt{3}}{2} = \sqrt{3}\,s$.The time interval between the balls striking the ground is $\Delta t = T_2 - T_1 = (\sqrt{3} - 1)\,s$.

Two balls are thrown simultaneously from the ground with the same velocity of $10\,m/s$ but at different angles of projection with the horizontal. Both balls fall at the same distance $5\sqrt{3}\,m$ from the point of projection. What is the time interval between the balls striking the ground?

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