Question

A particle of mass 1 kg has been thrown with initial
speed 20 m/s making an angle 60° with the
horizontal ground. The angular momentum of the
particle about point of projection when the projectile
is at the highest point is (g = 10 m/s)
(1) 150 kg m2/s (2) 300 kg m2/s 2
(3) Zero
(4) 100 kg m</s​

Answer 1

Answer:

L = 100 kg m/s

Explanation:

At the highest point, velocity = vcos60= 0.5V in horizontal direction

L= Angular momentum = Momentum × Perpendicular distance

Perpendicular distance is = V^2 / 4 g

Momentum =  mV/ 2 x V^2 / 4 g

As per given information we have;

V = 20

M = 1

G = 10

Putting the values we get;

L = (1 x 20 / 2) x (20^2 / 4 x 10)

L = 100 kg m/s