Question

How to find height of the particle at a give point and time in inclined plane motion?

Answer 1

Angle of projection=60

Angle of incline=30

Initial velocity=10m/s

 

Horizontal component of velocity=10cos60=5m/s

Vertical component of velocity=10sin60=5?3m/s

 

Now, the horizontal component of the velocity remains constant during the projectile motion.

 

Let us suppose the velocity of the particle is v when it is parallel to the inclined plane.

This means that at this instant this velocity is making an angle of 30 degrees with the horizontal.

 

Hence, at this instant,

vertical component of velocity=vcos30

It should be equal to the initial horizontal component

Hence, vcos30=5

or, v=5/cos30 = 10/?3m/s

 

Now, vertical velocity = vsin30= 10/2?3=5/?3m/s

 

Now for the vertical direction, the acceleration is g downwards, let us suppose time taken to reach this velocity is t, then for the vertical direction we can write

v=u-gt

or, 5/?3=5?3-10t

or, 10t = 5?3 - 5/?3= 10/?3

t= 1/?3

 

Hence time taken is 1/?3 seconds.

This is a easy method,
hope you understand...☺