Answer
The sled lands a distance of 25.6 meters from the foot of the cliff.
Work Step by Step
We can find the speed $v$ at the top of the cliff.
$K_2+U_2 = K_1+U_1$
$K_2 = K_1+0-U_2$
$\frac{1}{2}mv^2 = \frac{1}{2}mv_0^2-mgh$
$v^2 = v_0^2-2gh$
$v = \sqrt{v_0^2-2gh}$
$v = \sqrt{(22.5~m/s)^2-(2)(9.80~m/s^2)(11.0~m)}$
$v = 17.05~m/s$
We can find the time for the sled to fall a vertical distance of 11.0 meters.
$y = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{(2)(11.0~m)}{9.80~m/s^2}}$
$t = 1.50~s$
We can find the horizontal distance the sled travels in this time.
$d = vt = (17.05~m/s)(1.50~s)$
$d = 25.6~m$
The sled lands a distance of 25.6 meters from the foot of the cliff.
