question_answer 64)

                  A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle \[\theta \] with speed \[{{\upsilon }_{0}}\] and rebounds elastically (Fig.). Find (i) die distance along the plane where it will hit second time.

Answer:

                  Time of flight, \[T=\,\frac{2\,{{\upsilon }_{0}}\,\cos \,\theta }{g\,\cos \,\theta }=\,\frac{2{{\upsilon }_{0}}}{g}\] \ Distance travelled along the plane where it will hit second time, \[L=\,uT+\frac{1}{2}\,g\,\sin \,\theta \,\,({{T}^{2}})\] \[=\,({{\upsilon }_{0}}\,\sin \,\theta )\,\left( \frac{2{{\upsilon }_{0}}}{g} \right)\,\,+\frac{1}{2}\,g\,\sin \,\theta \,\,\left( \frac{4\upsilon _{0}^{2}}{{{g}^{2}}} \right)\]                 \[=\,\frac{2\upsilon _{0}^{2}\,\sin \,\theta }{g}+\,\frac{2\upsilon _{0}^{2}\,\sin \,\theta }{g}=\,\frac{4\upsilon _{0}^{2}\,\sin \,\theta }{g}\].