A particle of mass m is projected at t = 0 from a point O on the ground with a speed v 0 at an angle of 45 ° to the horizontal. The magnitude of the angular momentum of the particle about O at the time t = v 0 / g is close to [Take 2 = 1 . 4 ]

Step-by-step solution

Using 1st-equation of motion v = u + a t along horizontal direction, v x = v 0 / 2 + 0 × t = 0.7v as a = 0 and along vertical direction v y = v 0 / 2 - g v 0 / g = - 0.3v 0 So, the velocity of particle at time t = (v 0 /g) in vector form will be v → = i → 0.7v 0 - j → 0.3v 0 ....(i) Now from 2nd equation of motion s = ut + 1 2 at 2 along horizontal direction, x = v 0 2 × v 0 g = 0.7 v 0 2 g and along vertical direction, y = v 0 2 × v 0 g - 1 2 g v 0 g 2 = 0.2 v 0 2 g So the position vector r → at the t ( = v 0 /g)