A ball is projected from the bottom of an inclined plane of inclination ${30}^{\mathrm{o }}$, with a velocity of $\mathrm{30 }\mathrm{m }{\mathrm{s}}^{-1}$, at an angle of ${30}^o$ with the inclined plane. If  $g=\mathrm{10 }{\mathrm{ms}}^{-2}$, then the range of the ball on given inclined plane is

• A

  $12 \mathrm{m}$
• Correct answer
  B

  $60 \mathrm{m}$
• C

  $120 \mathrm{m}$
• D

  $600 \mathrm{m}$

Solution

Range on incline plane,
 $\mathit{R}=\frac{{2u}^2\mathit{sin}\left(\alpha -\beta \right)cos\alpha }{\mathit{g}{\mathit{cos}}^2\beta }$
Here,  $\beta ={30}^{\text{ο}}$ and $\alpha ={30}^{\text{ο}}+{30}^{\text{ο}}={60}^{\text{ο}}$

$\mathrm{ R}=\frac{2{\left(30\right)}^2\sin \left(30°\right)\cos \left(60°\right)}{10{\cos }^2\left(30°\right)}$
    $=\frac{2{\left(30\right)}^2\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)}{10\left(\frac{3}{4}\right)}$
    $=60\text{m}$

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