Question

Initial velocity with which a body is projected is $10m/s$ 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 $g=10m/s^2$ ]

• 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, $u=10m/s$
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
$R=\frac{2u^2}{g}\cdot \frac{\sin \theta \cos (\theta +\alpha )}{{\cos }^2\alpha }$
$=\frac{2\times 10^2}{10}\times \frac{\sin 30^{\circ }\cos \left(30^{\circ }+30^{\circ }\right)}{{\cos }^{2}30^{\circ }}$
$=20\times \frac{\frac{1}{2}\times \frac{1}{2}}{{\left(\frac{\sqrt{3}}{2}\right)}^2}=20\times \frac{\frac{1}{4}}{\frac{3}{4}}=\frac{20}{3}m$