Question

In Fig. the angle of inclination of the inclined plane is ${30}^0$. Find the horizontal velocity $V_0$ so that the particle hits the inclined plane perpendicularly.

• A. $V_0=\sqrt{\frac{2gH}{5}}$
• B. $V_0=\sqrt{\frac{2gH}{7}}$
• C. $V_0=\sqrt{\frac{gH}{5}}$
• D. $V_0=\sqrt{\frac{gH}{7}}$

Answer 1

The correct option is A $V_0=\sqrt{\frac{2gH}{5}}$

its is hitting perpendicularly means along the inclined plane its velocity is zero

so,along inclined plane

$V=U+at$

$0=V_{0}cos30-gsin30t$

$t=\frac{V_{0}cos30}{gsin30}=\frac{V_0\sqrt{3}}{g}$

along the direction perpendicular to the inclined plane

$S=Ut+\frac{1}{2}a{t}^2$

$\frac{H}{2}=V_{0}sin30\frac{V_0\sqrt{3}}{g}+\frac{1}{2}gcos30{\left(\frac{v_0\sqrt{3}}{g}\right)}^2$

$H=\frac{{V_0}^2}{g}[1+\frac{3}{2}]$

$V_0=\sqrt{\frac{2gH}{5}}$