Question

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.


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Answer 1

Answer:

The initial velocity of the projected object is given as:

v0=15iˆ+20jˆv0=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

​Using 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

​Therefore, the velocity of the particle after 2 s,

v=vxiˆ+vyjˆ=15iˆ+0.4jˆ

Explanation:

Answer 2

Answer:

The initial velocity of the projected object is given as:

v0=15iˆ+20jˆv0=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

​Using 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

​Therefore, the velocity of the particle after 2 s,

v=vxiˆ+vyjˆ=15iˆ+0.4jˆ

Explanation: