Answer
The man lands a horizontal distance of 4.9 meters from the edge of the ledge.
Work Step by Step
We can use conservation of energy to find the skier's speed at the bottom of the slope:
$\frac{1}{2}mv^2 = mgh$
$v = \sqrt{2gh}$
$v = \sqrt{(2)(9.80~m/s^2)(5.0~m)}$
$v = 9.90~m/s$
We can use conservation of momentum to find the speed after picking up the backpack:
$m_f~v_f = m_0~v_0$
$v_f = \frac{m_0~v_0}{m_f}$
$v_f = \frac{(65~kg)(9.90~m/s)}{65~kg+20~kg}$
$v_f = 7.63~m/s$
We can find the time it takes to fall 2.0 meters:
$y = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2y}{g}}$
$t = \sqrt{\frac{(2)(2.0~m)}{9.80~m/s^2}}$
$t = 0.639~s$
We can find the horizontal distance the skier (with the backpack) moves in this time:
$x = v_x~t = (7.63~m/s)(0.639~s) = 4.9~m$
The man lands a horizontal distance of 4.9 meters from the edge of the ledge.
