How far up the incline does the block travel before going back down?

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A small block travels up a frictionless incline that is at an angle of 30.0° above the horizontal. The block has speed 4.21 m/s at the bottom of the incline. Assume 

g = 9.80 m/s2.
 How far up the incline (measured parallel to the surface of the incline) does the block travel before it starts to slide back down? 
Oct 3rd, 2015

Thank you for the opportunity to help you with your question!

THE NET acceleration acting on the block is =  (- g Cos 60)

so, u = 4.21m/s

a = (- g Cos 60)  = -9.8 x 0.5 = -4.9m/s

v = 0 m/s

applying v^2 = u^2 + 2aS

where S is the distance covered.

on calculating S= 1.80 m

Please let me know if you need any clarification for the doubts. I'm always happy to answer your questions. Looking forward to help you again. thank you. :)
Sep 21st, 2015

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