An inclined plane is located at angle $\alpha =53°$ to the horizontal. There is a hole at point $\mathrm{B}$ in the inclined plane as shown in the figure. A particle is projected along the plane with speed $v_0$ at an angle $\beta =37°$ to the horizontal in such a way so that it gets into the hole. Neglect any type of friction. Find the speed $v_0$ (in $\mathrm{m} {\mathrm{s}}^{-1}$) if $h=1 \mathrm{m}$ and $l=8m$.

• Correct answer
  A

  9
• B

  3
• C

  6
• D

  12

Solution

$h=l\tan \beta -\frac{1}{2}\frac{g\sin \alpha l^2}{v_0^2{\cos }^2\beta }$
$v_0=\sqrt{\frac{g\sin \alpha l^2}{2{\cos }^2\beta \left(l\tan \beta -h\right)}}$
$=\sqrt{\frac{10\times \frac{4}{5}\times 8\times 8}{2\times \frac{4}{5}\times \frac{4}{5}\left(8\times \frac{3}{4}-1\right)}}=9 \mathrm{m} {\mathrm{s}}^{-1}$

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