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A particle is projected at an angle 60^@...

A particle is projected at an angle `60^@` with speed `10(sqrt3)m//s`, from the point A, as shown in the figure. At the same time the wedge is made to move with speed `10 (sqrt3) m//s` towards right as shown in the figure. Then the time after which particle will strike with wedge is

A

`2sec`

B

`2sqrt(3) sec`

C

`(4)/(sqrt(3)) sec`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
Suppose particle strikes wedge at height 'S' after time `t. S = 15 t - (1)/(2) 10 t_(2) = 15 t - 5 t_(2)`
During this time distance travelled by particle in horizontal direction `= 5 sqrt(3) t`. Also wedge has travelled extra distance

x= Total distance travelled by wedge in time
`t =10 sqrt(3)t.=5 sqrt(3) t + sqrt(3)(15-5t_(2))`
`sqrt(3) t.=5 sqrt(3) t+ sqrt(3) (15-5 t_(2))`
`rArr t =2 sec`
Alternative Sol.
(by Relative Motion)
`T = (2u sin 30^(@))/(g cos 30^(@)) = (2xx 10sqrt(3))/(10) xx "(1)/(sqrt(3))=2 sec`
`rArr t =2 sec`
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