Answer:
As
the block starts falling on the inclined plane, let at an instant, when the bolt
reaches at E, its velocity be v parallel to PQ. It then falls down due to
gravity pull. Taking vertical downward motion perpendicular to plane PQ of the
bolt, we have
u = 0, \[\text{a}\,\text{=}\,\text{gcos}\alpha \], \[s=l\], t = ?
As \[\text{s}=\text{ut}+\frac{1}{2}\text{a}{{\text{t}}^{\text{2}}}\]
\[\text{l}=0+\frac{1}{2}\text{g cos}\alpha
{{\text{t}}^{\text{2}}}\]
or \[\text{t}=\sqrt{\frac{2l}{g\cos \alpha }}\]
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