A particle of mass m is projected at an angle of `45^(@)` with the horizontal. At the highest point of its motion (h), what will be its angular momentum with respect to the point of projection?

Explanation

At any point, the horizontal component of velocity of the particle `= v_(x)` = vcos`45^(@)= (v)/(sqrt(2))`, the vertical velocity at the highest point =0
If the time taken by the particle to reach the highest point is t, then
0 = v`sin 45^(@)` - gt
or, `t=(v)/(sqrt(2 g))` [ initial vertical velocity = v sin`45^(@) ] " " cdots (1)`
If the maximum height attained is h, then
h = v sin`45^(@).t-(1)/(2)"gt"^(2)= (v)/(sqrt(2)).(v)/(sqrt(2)g)-(1)/(2)g.(v^(2))/(2g^(2))`
`=(v^(2))/(2 g)-(v^(2))/(4 g)=(v^(2))/(4 g)" "cdots(2)`
`:.` Angular momentum of the particle with respect to the point of projection
`=mvxxh=(mv)/(sqrt(2)).(v^(2))/(4 g)=(mv^(3))/(4 sqrt(2g))`
From equation (2) we get,
`v^(2)= 4 gh or, v = sqrt(gh)`
`:.` The angular momentum of the particle about the point of projection
`=(m)/(4sqrt(2g)).8.(gh)^(3//2)=mh sqrt(2gh)`