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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

`2 sec`

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=15t-(1)/(2)10t^(2)=15t`. During this time distance travelled by particle in horizontal direction `=5sqrt(3)t`. Also wedge has travelled extra distance

`x=(S)/(tan30^(@))=(15t-5t^(2))/(1//sqrt(3))`
Total distance travelled by wedge in time
`t=10sqrt(3)t.=5sqrt(3)+sqrt(3)(15-15t)`
`rArr t=2sec.`
Alternate Sol.
(by Relative Motion)

`T=(2usin30^(@))/(g cos30^(@))=(2xx10sqrt(3))/(10)xx(1)/(sqrt(3))=2sec.` `rArrt=2 sec.`
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