Question

A particle is projected from point P on inclined plane OA perpendicular to it with certain velocity v. It hits Banother inclined plane OB at point Q perpendicular toit. Point P and Q are at $h_1$ and $h_2$ height from ground $(\alpha >\beta )$

• A. $h_1=h_2$
• B. $h_1>h_2$
• C. $h_1<h_2$
• D. any of above depends on speed of projection

Answer 1

The correct option is B $h_1>h_2$

$v^{\prime }$ (perpendicular to OB) $=-v\sin (0^o-\alpha -\beta )+g\cos \beta t$ .......(i)

Parallel to OB, O=$v\cos ({90}^o-\alpha -\beta )-g\sin \beta t$

On solving, $v^{\prime }=\frac{v\sin \alpha }{\sin \beta }$

for $\alpha >\beta ,v^{\prime }>v$

$\therefore h_2<h_1$