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?

A particle of mass m is projected at an angle of 45^(@) with the horiz

1.	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?
Answer» Solution :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 ish, 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)`