Question

An object is launched from a cliff $20m$ above the ground at an angle of ${30}^o$ above the horizontal with an initial speed of $30m/s$. How far does the object travel before landing on the ground? (in metre)

• A. $20$
• B. $20\sqrt{3}$
• C. $60$
• D. $60\sqrt{3}$

Answer 1

The correct option is D $60\sqrt{3}$

Height of the cliff $h_1=20m$

Velocity of projectile $V=30m/s$

Initial upward velocity $v_y=V\sin \theta =30\times sin{30}^{{}^{\circ }}=15m/s.$

Horizontal velocity, $v_x=V\cos \theta =30\cos {30}^{o}m/s$

Time taken to reach maximum height, $t_1=\frac{V_y}{g}=\frac{15}{10}=1.5sec.$

Height above the cliff $h_2=\frac{v_y^2}{2g}=\frac{{15}^2}{2\times 10}=11.25m$

Therefore, the maximum height, $h_{max}=h_1+h_2=20+11.25=31.25m.$

Time taken to free fall from max height, $t_2=\sqrt{\frac{2h_{max}}{g}}=\sqrt{\frac{2\times 31.25}{g}}=2.5sec.$

Thus, the total time taken during the entire flight $t_{total}=t_1+t_2=1.5+2.5=4\sec$

Total horizontal distance covered $R=v_x{t}_{total}=30\times cos{30}^{{}^{\circ }}\times 4=60\sqrt{3}$