Answer:
[tex]h=3.7 m[/tex]
Explanation:
Apply the theorem of the work and conservation of energy the motion have two steps the first without picks up the woman
[tex]W_g=K_f-K_i[/tex]
[tex]K_i=0[/tex]
[tex]K_f=\frac{1}{2}*m_1*v_1^2=\frac{1}{2}75kg*v_1^2[/tex]
[tex]W_g=m*g*R[/tex]
Solve to v'
[tex]m*g*R=\frac{1}{2}*m*v^2[/tex]
[tex]v^2=2*g*R[/tex]
[tex]v=\sqrt{2*9.8m/s^2*10.8m}=\sqrt{211.68m^2/s^2}[/tex]
[tex]v=14.54 m/s[/tex]
The momentum is conserved so the momentum as a perfectly inelastic collision give that the woman began in rest so
[tex]p_1-p_2=pf[/tex]
[tex]m*v_1+m_2*v_2=(m_1+m_2)v_t[/tex]
[tex]v_2=0[/tex]
[tex]75.0kg*14.5 m/s=(75.0kg+51.5kg)*v_t[/tex]
[tex]v_t=\frac{1091 kg*m/s}{75.0kg+51.5kg}=8.6m/s[/tex]
Finally the kinetic energy is the same of the potential energy so can find the high the reach in their upward swing
[tex]K_t=E_p[/tex]
[tex]\frac{1}{2}*(m_1+m_2)*v_t^2=(m_1+m_2)*g*h[/tex]
Solve to h'
[tex]h=\frac{v_t^2}{2*g}=\frac{(8.6m/s)^2}{2*9.8m/s^2}=\frac{73.96 m^2/s^2}{19.6m/s^2}[/tex]
[tex]h=3.7 m[/tex]
