Question

Initial velocity with which a body is projected is $10 m / s$ 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 value of $g =10 \,m / s ^{2}$ ]

• A. $\frac{15 \sqrt{3}}{2} m$
• B. $\frac{40}{3} m$
• C. $5 \sqrt{3} m$
• D. $\frac{20}{3} m$

Answer 1

Answer: (D)

Initial velocity of projectile, $u=10 \,m / s$
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}} $
$=20 \times \frac{\frac{1}{2} \times \frac{1}{2}}{\left(\frac{\sqrt{3}}{2}\right)^{2}}=20 \times \frac{\frac{1}{4}}{\frac{3}{4}}=\frac{20}{3} m$