InterviewSolution
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InterviewSolution
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Two inclined planes OA and OB having inclination (with horizontal) `30^(@) and 60^(@)`, respectively, intersect each other at O as shown in figure. A particle is projected from point P with velocity `u = 10sqrt3 ms^(-1)` along a direction perpendicular to plane OA. If the particle strikes plane OB perpendicularly at Q, calculate The vertical height h of P from O,A. The time of flight `2s`B. The velocity with which the particle strikes the plane `OB=10 m//s`C. The height of the point `P` from point `O` is `5m`D. The distance `PQ=20m` |
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Answer» Correct Answer - A::B::C::D Consider the motion of particle along the axes shown in figures.We have `u_(x)=u, a_(x)=-g sin 60^(@)` `u_(y)=0, a_(y)=-g cos 60^(@)` (a) As the particle strikes the plane `OB` perpendicularly, `:.v_(x)=0` as `v_(x)=u_(x)+a_(x)t` or `0=u-g sin 60^(@)t rArr t=u/(g sin 60^(@))=(10sqrt3)/(10xxsqrt3/2)=2s` (b) Initial velocity along `y`-axis is zero.The velocity along `y`-axis after `2 s ,v_(y)=u_(y)+a_(y)t` `=0-g cos 60^(@) xx2=-10xx1/2xx2=-10m//s` (c) We have , `v_(x)^(2)=u_(x)^(2)+2a_(x)s` Since `v_(x)=0` and `a_(x)=g sin 60^(@),u=10sqrt3m//s` `:.0=(10sqrt3)^(2)-2xxg sin 60^(@)xx(OQ)` `OQ=(10^(2)xx3)/(2xx10xxsqrt3/2)=10sqrt3m` Distance`PO=0+1/2g sin 30^(@)xx(2)^(2)` `=1/2xx10xx1/2xx4=10m` Therefore height `h` of point `P`, `h=PQ sin 30^(@)=10xx1/2=5m` (d) Distance `PQ=sqrt(PO^(2)+OQ^(2))` `=PQ=sqrt((10)^(2)+(10sqrt3)^(2))=20m` |
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