## College Physics (4th Edition)

The second ledge should be placed at a height of $1.27~meters$ above the ground.
We can use conservation of energy to find Jones' speed $v_0$ just before the collision: $KE = U_g$ $\frac{1}{2}m_0v_0^2 = m_0gh$ $v_0 = \sqrt{2gh}$ $v_0 = \sqrt{(2)(9.80~m/s^2)(3.70~m})$ $v_0 = 8.516~m/s$ We can use conservation of momentum to find their speed $v_f$ just after the collision: $m_f~v_f= m_0~v_0$ $v_f= \frac{m_0~v_0}{m_f}$ $v_f= \frac{(78.0~kg)~(8.516~m/s)}{78.0~kg+55.0~kg}$ $v_f = 4.99~m/s$ We can use conservation of energy to find the height that they reach: $U_g = KE$ $m_fgh = \frac{1}{2}m_fv_f^2$ $h = \frac{(v_f)^2}{2g}$ $h = \frac{(4.99~m/s)^2}{(2)(9.80~m/s^2)}$ $h = 1.27~m$ The second ledge should be placed at a height of $~1.27~meters~$ above the ground.