Question

Determine the horizontal velocity $v_0$ with which a stone must be projected horizontally from a point P, so that it may hit the inclined plane perpendicularly. The inclination of the plane with the horizontal is $\theta$ and point P is at a height h above the foot of the incline, as shown in the figure.

• A. $v_0=\sqrt{\frac{4gh}{2+{\cot }^2\theta }}$
• B. $v_0=\sqrt{\frac{2gh}{2+{\cot }^2\theta }}$
• C. $v_0=\sqrt{\frac{2gh}{1+{\cot }^2\theta }}$
• D. $v_0=\sqrt{\frac{4gh}{1+{\cot }^2\theta }}$

Answer 1

The correct option is B $v_0=\sqrt{\frac{2gh}{2+{\cot }^2\theta }}$

X direction :

Using $V=u+at$ $⟹$ $V_x=v_{o}cos\theta -\left(gsin\theta \right)t$

As the stone hits the inclined plane perpendicularly, thus the stone has the final velocity in y direction only i.e $V_x=0$

$\therefore$ $0=v_{o}cos\theta -gsin\theta t$ $⟹t=\frac{v_o}{gtan\theta }$ ............(1)

y direction :

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

$\therefore$ $hcos\theta =\left(v_{o}sin\theta \right)t+\frac{1}{2}\times \left(gcos\theta \right)t^2$

OR $hcos\theta =\left(v_{o}sin\theta \right)\times \frac{v_o}{gtan\theta }+\frac{gcos\theta }{2}\times \frac{v_o^2}{g^{2}ta{n}^2\theta }$

OR $h=\frac{v_o^2}{g}+\frac{v_o^{2}co{t}^2\theta }{2g}$

OR $v_o^2=\frac{2gh}{2+co{t}^2\theta }$

$⟹$ $v_o=\sqrt{\frac{2gh}{2+co{t}^2\theta }}$