The velocity of a projectile at the initial point A is (2 i + 3 j) m/s. Its velocity (in m/s) at point B is [figure]
- 2 i - 3 j
- 2 i + 3 j
- -2 i - 3 j
- -2 i + 3 j
Answer
(1) 2 i - 3 j
Solution
Horizontal component of velocity remains constant: vx = 2 m/s
B is at the same level as A, so the vertical component has the same magnitude but is now downward: vy = -3 m/s
vB = 2 i - 3 j
Answer: (1)