Answer to Question #280148 in Physics for dfghj

Question

An 70 kilogram guy descends from a height of 10.0 m by holding onto a rope that goes through a frictionless pulley to a 55 kg sandbag. What is the speed at which the man hits the ground if he starts from rest?

Answer 1

$T-m_1g=-m_1a$

$T-m_2g=m_2a\to$

$a=\frac{m_1-m_2}{m_1+m_2}g=\frac{70-55}{70+55}\cdot9.8=1.18\ (m/s)$

$h=\frac{v^2}{2a}\to v=\sqrt{2ah}=\sqrt{2\cdot1.18\cdot10}=4.86\ (m/s)$ . Answer