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A particle of mass m is projected up fro...


A particle of mass `m` is projected up from the bottom of an inclined plane with initial velocity `v_(0)` at angle `45^(@)` with an inclined plane of inclination `30^(@)` as shown in figure. At the same time a small block of same mass `m` is released from rest at a height `h`, the particle hits the block at some point on the inclined plane. Neglect friction at inclined.
Q. The value of height h is

A

`(v_(0)^(2))/(g)`

B

`(sqrt(3)v_(0)^(2))/(g)`

C

`(2v_(0)^(2))/(sqrt(3)g)`

D

`(v_(0)^(2))/(sqrt(3)g)`

Text Solution

Verified by Experts

The correct Answer is:
D

`T=(2V_(0)sin45^(@))/(2gcos30)=(2sqrt(2)V_(0))/(sqrt(3)g)`
since `x_(1)+x_(2)=2h`
`implies((V_(0))/(sqrt(2))(2sqrt(2)v_(0))/(sqrt(3)g)-(1)/(2)gsin30^(@)xx(8v_(0)^(2))/(3g^(2)))+((1)/(2)(g)/(2)(8v_(0)^(2))/(3g))=2h`
`h=(v_(0)^(2))/(sqrt(3)g)`
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