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A particle is projected from ground with...

A particle is projected from ground with velocity `40sqrt2m` at `45^(@)`.At time `t=2s`

A

displacement of particles is `100 m`

B

vertcal component of velocity is `20 m//s`

C

velocity makes an angle of `tan^(-1)(2)` with vertical

D

particle is at height of `60 m` from ground

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D

`u_(x)=40m//s,u_(y)=40m//s`,At `t=2s`.
`v_(x)=40m//s,v_(y)=40m-10xx2=20m//s`.
`x=v_(x)t=80m`
`y=u_(x)t-1/2"gt"^(2)=60m`
`:.s=sqrt(x^(2)+y^(2))=100m`
`theta=tan^(-1)(v_(x)/v_(y))=tan ^(-1)(2)`
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