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A particle is projected at time t=0 from...

A particle is projected at time t=0 from a point P on the ground with a speed `v_0,` at an angle of `45^@` to the horizontal. Find the magnitude and direction of the angular momentum of the particle about P at tiem `t= v_0//g`

Text Solution

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Velocity at any time 't' is `vec(v) =v_(x)hati+v_(y)jhatj`

Position vector of projectile at time 't' is `vec(r)=x hati+yhatj` since `vec(L)=m(hatrxxhatv)`
`vec(L)=m[(xhati+yhatj)xx(v_(x)hati+v_(y)hatj)]`
`vec(L)=m[xv_(x)hatk-yv_(x)hatk], bar(L)=mbar(K)[xv_(y)-yv_(A)]`
where `v_(x)=u/(sqrt(2))` and `x=(u^(2))/(sqrt(2)g)`
`:. x=(u cos 45^(@))t=u/(sqrt(2))((u)/(g))`
`v_(y)=u sin 45^(@)-"gt"=(1-sqrt(2))/(sqrt(2))v_(0)`
`y=u sin 45^(@)t-1/2"gt"^(2)=(y^(2))/(2g)(sqrt(2)-1), vec(L)=(-hatkmu^(3))/(2sqrt(2)g)`
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