Question

A particle of mass m is projected with velocity v making an angle of 45 with the horizontal

Answer 1

At the highest point the vertical distance of the mass is d = [v^2sin^2(45)]/2g=v^2/(4g) At the highest point it has only horizontal component of velocity so this momentum will me =mvcos45=mv/✓2 Hence angular momentum will be = v^2/(4g)*mv/✓2=mv^2/(4✓2g)Read more on Sarthaks.com - https://www.sarthaks.com/204220/a-particle-of-mass-m-is-projected-with-velocity-v-making-an-angle-of-45-with-the-horizontal