An object is projected with velocity v0=15i^ +20j^. Considering x along horizontal axis and y along vertical axis. Find its velocity after 2 seonds.

The initial velocity of the projected object is given as:
v0=15i^+20j^
There is acceleration in the vertical direction only. There is no acceleration in the horizontal direction, so the velocity in the horizontal direction will remain the same.
Therefore, the velocity in the horizontal direction after t = 2 s, vx = 15 m/s
‚ÄčUsing equation of motion, let us find the velocity of the object in vertical direction after t = 2 s:
vy = uy -gt
vy = 20 - 9.8×2 = 20-19.6 = 0.4 m/s
‚ÄčTherefore, the velocity of the particle after 2 s,
v=vxi^+vyj^=15i^+0.4j^
 

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