A particle is projected at an angle of 30^ from the horizontal with...

Given, angle of projection, $\theta=30^{\circ}$ Initial speed of particle, $\mathrm{v}=10 \mathrm{~m} / \mathrm{s}$ Time of flights, range and velocity of impact on the surface of earth are T, R and V respectively $\therefore \quad \mathrm{T}=\frac{2 \mathrm{v} \sin \theta}{\mathrm{g}}, \mathrm{R}=\frac{\mathrm{v}^2 \sin 2 \theta}{\mathrm{g}}$ Gravitational acceleration on moon's surface is equal to $\frac{1}{6}$ part of gravitational acceleration on the earth's surface. Hence, $$\begin{aligned} & g_m=\frac{g_e}{6} \\ & T_m=\frac{2 v \sin \theta}{\frac{g}{6}}=\frac{6 \times 2 v \sin \theta}{g}=6 \mathrm{~T} \\ & R_m=\frac{v^2 \sin 2 \theta}{\frac{g}{6}}=\frac{6 v^2 \sin 2 \theta}{g}=6 R \end{aligned}$$ Velocity of impact will be same at both places of moon and earth. $\therefore \quad \mathrm{V}_{\mathrm{m}}=\mathrm{V}$