This is a question of Kinametics : Projectile motion on an inclined plane: A particle is thrown at time t=0, with a velocity of 10 m/s at an angle of 60 degree with the horizontal, from a point on an incline plane, making an angle of 30 degree with the horizontal. Find the time when the velocity of the projectile becomes parallel to the incline.

Asked by prernamehta | 1st Apr, 2012, 12:04: PM

Expert Answer:

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.
 
Regards
Toppers Team

Answered by  | 9th Apr, 2012, 04:58: PM

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