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A particle is projected with a velocity ...

A particle is projected with a velocity of `10sqrt2 m//s` at an angle of `45^(@)` with the horizontal. Find the interval between the moments when speed is `sqrt125 m//s(g=10m//s^(2))`

Text Solution

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`v=sqrt(125)`m/s
`u_(x)=10sqrt(2) cos45^(@)=10" m"//s`,
`u_(y)=10sqrt(2) sin 45^(@)=10" m//s`
`v^(2)=v_(x)^(2)+v_(y)^(2)`
`125=100+v_(y)^(2) rArr v_(y)=5" m"//"s"" "(:. v_(x)=u_(x))`
The required time interval is
`Deltat=(2v_(y))/(g)=(2xx5)/(10)=1s`
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