Question

The velocity of a projectile at the initial point A is $(2\hat{i}+3\hat{j})m/s$. It's velocity (in m/s) point B is

Answer 1

Initial velocity,$v=(2\hat{i}+3\hat{j})m{s}^{-1}$

$v=v_x\hat{i}+v_y\hat{j}$

Horizontal velocity remain constant during flight.

At point B, horizontal velocity, $v_b=2\hat{i}m{s}^{-1}$

Hence, velocity at point B is $2\hat{i}m{s}^{-1}$