Welcome to the forum.

I like to start with a drawing with the forces shown in red. (see below)

I've got the force due to gravity as mg

the normal reaction of the slope as R

and I've put in a frictional force F even though it isn't needed for the first part.

So with F = 0

the acceleration down the slope is a = mg times 3/5

you know the initial velocity u = 28

it will stop going up when its velocity v = 0

so you can get the distance, s, using

and for the time, t

On the way back u will now be zero, 'a' is reversed, s is unchanged so v is easy to get.

When there's a frictional force you'll need to resolve along the slope and perpendicular to it

That's for the travel up the slope.

On the way back the friction will oppose the motion, so will be up the slope.

Bob

]]>thank you very much in advance.

]]>