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

Dear Pranay

max height ,H= v²sin²θ/2g

velocity at max height = u_x= vcosθ

angular momentum= mu_xH

and putting θ =45⁰

angular momentum = mv³/g4√2

All the best.

AKASH GOYAL

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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)