Initial velocity with which a body is projected is $10 \mathrm{~m} / \mathrm{s}$ from the base of an inclined plane as shown in the given figure. If the angle of projection is $60^{\circ}$ with the horizontal, then the range $R$ is [take, $g=10 \mathrm{~m} / \mathrm{s}^2$ ]

• A

  $\frac{15 \sqrt{3}}{2} \mathrm{~m}$
• B

  $\frac{40}{3} m$
• C

  $5 \sqrt{3} \mathrm{~m}$
• Correct answer
  D

  $\frac{20}{3} \mathrm{~m}$

Solution

Initial velocity of projectile, $u=10 \mathrm{~m} / \mathrm{s}$ Angle of projection from inclined plane, $\theta=60^{\circ}-30^{\circ}=30^{\circ}$ Here, $\alpha=30^{\circ}$
Range over an inclined plane is given as $R=\frac{2 u^2}{g} \cdot \frac{\sin \theta \cos (\theta+\alpha)}{\cos ^2 \alpha}$ $=\frac{2 \times 10^2}{10} \times \frac{\sin 30^{\circ} \cos \left(30^{\circ}+30^{\circ}\right)}{\cos ^2 30^{\circ}}$ $=20 \times \frac{\frac{1}{2} \times \frac{1}{2}}{\left(\frac{\sqrt{3}}{2}\right)^2}=20 \times \frac{\frac{1}{4}}{\frac{3}{4}}=\frac{20}{3} \mathrm{~m}$

Hence, the correct answer is option: 'D'.