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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


Option: 2

 60 \sqrt{3}

Option: 3

 60 \sqrt{2}


Option: 4


Answers (1)


 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}

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