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`        a particle of mass m is projected with a velocity v making an angle of 45 degrees with the horizontal. the magnitude of the angular momentum of the projectile about the point pf projection when the particle is at its max height `
7 years ago

419 Points
```										Dear Pranay
max height ,H= v2sin2θ/2g
velocity at max height = ux= vcosθ
angular momentum= muxH
and putting θ =450
angular momentum = mv3/g4√2

All the best.
AKASH GOYAL

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```
7 years ago
510 Points
```										if the point of projection is taken as origen then

coordinates of maximum height (H,R/2)       or r =Hi + (R/2)j       in vector  form
R is horizontal  range  and H is maximum height ....

at maximum height there is only horizontal component of velocity no vertical component exists ...
so , V= vcos@i  ......
linear momentam = mvcos@i                     ( @ is the angle of projection)
we know that angular momentam(L) about point of projection = r*p                (* is cross product)
L  =(Hi + (R/2)j)*(vi)
L  =Rv/2 (-K)
L=v3 /2g (-k)                                       ( R =v2 /g at @=45 degree)
```
6 years ago
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