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The velocity of a projectile at the initial point A is $(2\hat i + 3\hat j)$ m/s. Its velocity (in m/s) at point B is:
A
$2\hat i + 3\hat j$
B
$-2\hat i - 3\hat j$
C
$-2\hat i + 3\hat j$
D
$2\hat i - 3\hat j$
Detailed Solution
B is at the same level as A. The horizontal component of velocity stays the same and the vertical component is reversed.
So the velocity at B is $2\hat i - 3\hat j$ m/s.
So the velocity at B is $2\hat i - 3\hat j$ m/s.