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  • A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle \theta with speed v_{o} and rebounds elastically (Fig 4.7). Find the distance along the plane where it will hit second time.

A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle \theta with speed v_{o} and rebounds elastically (Fig 4.7). Find the distance along the plane where it will hit a second time.



 

Answers (1)

The particle rebounds from P. when it strikes plane inclined at v0 speed. Hence the speed of particle after it rebounds from P will be v0 We assume the new axis X’OX and YOY’ axis at P as origin ‘O’.  The components of g and v0 in the new OX and OY axis are:

Focusing on the motion of the particle from O to A,

s=ut+\frac{1}{2}gt^{2}

Here, t=T which is the time of flight

0=T[v\; \cos\; \theta -\frac{1}{2}g\; \sin\; \theta \; T]

So, either T=0\; or \left [ v\; \cos\; \theta -\frac{1}{2}g\; \sin\; \theta \; T \right ]=0

S_{x}=u_{x}t+\frac{1}{2}a_{x}t^{2}

L=[\frac{2v}{g}]v\; \sin \theta +\frac{1}{2}g\; \sin\; \theta [\frac{2v}{g}]^{2}

L=\frac{4v^{2}}{g}.\sin \theta

Posted by

infoexpert23

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