Question

In figure the angle of inclination of the inclined plane is ${30}^{\circ }$. 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 $V_0=\sqrt{\frac{2gH}{5}}$


On taking components of initial velocity $V_0$ along x and y direction (i.e. along the inclined plane and perpendicular to the inclined plane), we get
$V_x=V_0\cos {30}^{\circ }$and
$V_y=-V_0\sin {30}^{\circ }$
Using first equation of motion along x-axis, we get
$V_x=u_x-g\sin {30}^{\circ }t$
As the final velocity is perpendicular to the inclined plane
$\Rightarrow V_x=0$
$0=V_0\cos {30}^{\circ }-g\sin {30}^{\circ }t\Rightarrow t=\frac{\sqrt{3}{V}_0}{g}$

Using second equation of motion along y-axis, we get
$y=u_{y}t-\frac{1}{2}g\cos \theta t^2\Rightarrow -H\sin {60}^{\circ }=-V_0\sin {30}^{\circ }t-\frac{1}{2}g\cos {30}^{\circ }{t}^2-\frac{\sqrt{3}H}{2}=-\frac{V_0}{2}\times \frac{\sqrt{3}{V}_0}{g}-\frac{\sqrt{3}g}{4}\times \frac{3V_0^2}{g^2}\Rightarrow H=\frac{V_0^2}{g}+\frac{3V_0^2}{2g}H=\frac{5V_0^2}{2g}\Rightarrow V_0=\sqrt{\frac{2gH}{5}}$