An object of mass 'm' is projected from origin in a vertical xy plane at an angle 45 with the x-axis with an initial velocity v0. The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [ g is acceleration due to gravity]
- mv032 2 g along negative z-axis
- mv032 2 g along positive z-axis
- mv034 2 g along positive z-axis
- mv034 2 g along negative z-axis
Answer
(4) mv034 2 g along negative z-axis
Solution
At the top v= v0 2 i, at height H= v024g above O.
[figure]
L=m v H=m v0 2 v024g= mv034 2 g
Velocity to the right and above O clockwise along - z
Answer: (4)