Question

An inclined plane makes as angle ${\theta }_0={30}^{\circ }$ with the horizontal. A particle is projected from this plane with a speed of $5\text{m/s}$ at an angle of elevation $\beta ={30}^{\circ }$ with the horizontal as shown in the figure.


Find the range of the particle on the plane when it strikes the plane. (Assume the incline is long enough and $g=10{\text{m/s}}^2$)

• A. $3.5\text{m}$
• B. $4\text{m}$
• C. $4.2\text{m}$
• D. $5\text{m}$

Answer 1

The correct option is D $5\text{m}$

Angle of inclination with the inclined surface, $\alpha ={\theta }_0+\beta ={60}^{\circ }$
Range on incline, $R=\frac{u^2}{g{\cos }^2{\theta }_0}\left[\sin \right(2\alpha -{\theta }_0)+\sin {\theta }_0]$
$=\frac{5^2}{10{\cos }^2{30}^{\circ }}\left[\sin \right(2\times {60}^{\circ }-{30}^{\circ })+\sin {30}^{\circ }]$
$=\frac{25}{10\times {\left(\frac{\sqrt{3}}{2}\right)}^2}[\sin {90}^{\circ }+\frac{1}{2}]$
$=\frac{25\times 4}{10\times 3}\times \frac{3}{2}$
= $5\text{m}$