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A particle is projected from the ground ...

A particle is projected from the ground . One second after the projection, the particle is found to be moving at an angle `45^(@)` with the horizontal . Speed of the particle becomes minimum two seconds after the projection . If angle of projection is `tan^(-1)(n)`, then what is the value of n ?

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The correct Answer is:
2
[2]: Let particle be projected with a speed u at an angle `theta` with the horizontal . Speed of the particle becomes minimum at the highest point . Time to reach the highest point can be written as follows :
`t=(u sin theta)/(g) rArr 2=(u sin theta)/(10) rArr u sin theta=20m//s`
Component of velocity after one second can be calculated as follows:
`u_(y)=u sin theta-"gt"=20-10xx1=10m//s`.
`tan 45^(@)=(u_(y))/(u_(x))=1 rArru_(x)=v_(y)=10m//s`
We know that x-component of velocity remains constant, hence
`u_(x)=u cos theta =10m//s`
Finally we have
`u sin theta =20, ucos theta=10`
On dividing we get
`tan theta =2`
`rArr theta tan^(-1)(2)`
Hence n=2
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