Question

derive the expression for final speed of a particle moving in an inclined plane​

Answer 1

Explanation:

final point /time taken to reach at that point

Answer 2

Answer:

$v=\sqrt{u^2-2gh}\ m/s$

Explanation:

Let

Mass of particle = m

Initial velocity = u m/s

Final velocity = v

Height of incline surface = h

We know that from work power energy theorem

Work due to all work = Change in kinetic energy

So From particle move from point A to B then

Work done to gravity = - m g h

Change in kinetic energy

ΔK= 1/2 m v² -  1/2 m u²

So

- m g h = 1/2 m v² -  1/2 m u²

-2 g h =   v² -    u²

 v² = u²- 2 g h

$v=\sqrt{u^2-2gh}\ m/s$