Why is acceleration g sin(10) on an inclined plane?

[HT3]

Homework Statement

A skier is gliding along at 3m/s on horizontal frictionless snow. He suddenly starts down a 10 degree incline. His speed at the bottom is 15 m/s.
a) what is the length of the incline?
b) how long does it take for him to reach the bottom?
we know:
Vo=3m/s
Vf=15m/s
t=?
x=?
a=? [gsin10 -- but why?]

Homework Equations

V^2=Vo^2 +2ax

The Attempt at a Solution

okay...i don't understand why a=gsin(10) [g=9.8]
because from the triangle i draw ..g is not the hypotenuse? I am confused to see WHY a=9.8sin(10).

http://img125.imageshack.us/img125/6060/53872528yt8.jpg

 

[cristo]

I don't think it is!

You're correct that g is not the hypotenuse. g acts vertically downwards, and so if you want the component of the acceleration due to gravity that acts down the slope you need to resolve this into components parallel and perpendicular to the slope. To do this, g is the vertical side of the triangle you've drawn, and you want to find the hypotneuse, x say. Hence, sin(10)=g/x and so x=g/sin(10)

 

[HT3]

thanks for replying so fast cristo!
okay so if i do that:
sin(10)=g/a and so a=g/sin(10)

a = 9.8/(sin10)
a=56.4m/s^2?
that just seems way out of wack...its incorrect...according to my solutions page it says a=gsin10. I just cannot seem to figure how that is visually according to what i see on the triangle. Any ideas?

 

[Hootenanny]

It is incorrect, since for it to be physically meaningful $a\leq g$. I think the question is defining the angle from the vertical (10 degrees from the horizontal is a pretty small incline for a ski slope).

 

[HT3]

yes...a bunny slope i suppose?
but even so
why is:
a=gsin(theta) --> from the vertical: a=gsin(80) frm the horizontal a=gsin(10)?
as u said it makes sense for a< or equal to g..but how is it that we get that as the trigonometric function we use...i just don't see the triangle...hopefully that makes sense...if someone could draw it out for me how we get a=9.8sin(10)=1.7m/s^2