An object is launched from a cliff  $20 \mathrm{~m}$ above the ground at an angle of $30^{\circ}$ above the horizontal with an initial speed of $30 \mathrm{~m} / \mathrm{s}$. How far does the object travel before landing on the ground? (in metres)Option: 1  20Option: 2  $60 \sqrt{3}$Option: 3  $60 \sqrt{2}$  Option: 4 40

Height of the cliff, $h_1=20 \mathrm{~m}$
Velocity of projectile ,$\mathrm{V}=30 \mathrm{~m} / \mathrm{s}$
Initial upward velocity,  $\mathrm{v}_{\mathrm{y}}=\mathrm{V} \sin \theta=30 \times \sin 30^{\circ}=15 \mathrm{~m} / \mathrm{s}$.
Horizontal velocity, $\mathrm{v}_{\mathrm{x}}=\mathrm{V} \cos \theta=30 \cos 30^{\circ} \mathrm{m} / \mathrm{s}$
Time taken to reach maximum height, $\mathrm{t}_1=\frac{\mathrm{V}_{\mathrm{y}}}{\mathrm{g}}=\frac{15}{10}=1.5 \mathrm{sec}$.
Height above the cliff, $\mathrm{h}_2=\frac{\mathrm{v}_{\mathrm{y}}^2}{2 \mathrm{~g}}=\frac{15^2}{2 \times 10}=11.25 \mathrm{~m}$

Therefore, the maximum height, $\mathrm{h}_{\max }=\mathrm{h}_1+\mathrm{h}_2=20+11.25=31.25 \mathrm{~m}$.
Time taken to free fall from max height, $\mathrm{t}_2=\sqrt{\frac{2 \mathrm{~h}_{\max }}{\mathrm{g}}}=\sqrt{\frac{2 \times 31.25}{\mathrm{~g}}}=2.5 \mathrm{sec}$.
Thus, the total time taken during the entire flight $t_{\text {total }}=t_1+t_2=1.5+2.5=4 \mathrm{sec}$
The total horizontal distance covered $\mathrm{R}=\mathrm{v}_{\mathrm{x}} \mathrm{t}_{\text {total }}=30 \times \cos 30^{\circ} \times 4=60 \sqrt{3}$