# Please check if I did it right?

1. Feb 24, 2008

### yaho8888

1. The problem statement, all variables and given/known data
A skier starts from rest at the top of a 26 degree slope 1.3Km long.
how long does it take to reach the bottom? (no friction)

The attempt at a solution
m=mass
m*g*sin(24) = ma(at slope direction)
divide both side by m
we got
g*sin(24)=a

then use
v^2(final) = V^2(initial)+2(9.8m/s^2)(sin(24))(1300m)
V^2(final) = 0+2(9.8m/s^2)(sin(24))(1300m) we know that initial v is zero.
we get V from the equation above.
then use
v= v(initial)+at
v=0+(9.8m/s^2)sin(24)t
t=V/9.8sin(24).
done!

2. Feb 24, 2008

### Mindscrape

I suppose it would be okay, but why not just use

$$s(t) = s_0 + v_0t +\frac{1}{2}at^2$$

and solve for t?

Did you mean a 24 degree slope, or are you putting in the wrong angle?