Question

A particle of mass m is projected from the ground with an initial speed u at an angle alpha. Find the magnitude of its angular momentum at the highest point of its trajector about the point of projection.

Answer 1

Answer:

i hope it helps u....

Explanation:

Refer figure, vertical distance, of particle from point of origin, when at highest point of its trajectory is H=(usinα)  

2

/2g.

The velocity at highest point is v  

h

​  

=ucosα.

So the angular momentum is mv  

h

​  

×H=mu  

3

cosαsin  

2

α/2g

so, x=2

Answer 2

Angular momentum at the highest point of its trajectory about the point of projection is m*u^3$cos^{2}$(∝)*sin(∝)/g Kg-meter ^2 per second

1. Here the velocity of the particle at the maximum height is ucos(∝).

2. The horizontal distance or range at that time =  ucos(∝)*t

                                                                                 = ucos(∝)*usin(∝)/g

                                                                                 =u^2*sin(∝)*cos(∝)/g

3. Therefore the angular momentum at the highest point of its trajectory about the point of projection = m*v*r

                                               =m*u^2*sin(∝)*cos(∝)/g*ucos(∝).

                                               = m*u^3$cos^{2}$(∝)*sin(∝)/g

where m= mass of the particle.