Question

A block is projected up a frictionless inclined plane with a
speed Vq. The angle of incline is 6 . (a) How far up the plane
does it go? (b) How long does it take to get there? (c) What is
its speed when it gets back to the bottom? Find numerical
answers for 0 = 35® and Vq= 8.2 ft/s

Answer 1

Answer:

use Motion Formulas

by v^2=u^2-2as

a=-g sin tita (if tita is the angle Inclined to horizontal)

now...0=vq^2-2g sin tita S

then S=vq^2/2g sin tita

v=u+at

0=vq-g sin tita t

So ..t =vq/g sin tita

the speed is the same speed vq come to the bottom.because it is frictionelless