Answer:

Given frictionless, massless pulley

The masses were released from rest

Height from which $60N$ mass falls is $h=2m$

Let the tension in the string be T(N) and the accelerations of the masses of $60N (w_1)$ and $30N (w_2)$ be $a_1(m/s^2)$ and $a_2(m/s^2) $ respectively and $g$ be acceleration due to gravity As the masses were directly connected we know that they have the same accelerations So $a_1=a_2$ To find the speed $v_1(m/s)$ at which $60N$ mass will hit the ground We know newton’s second law of motion $F=ma$ From the free body diagram of 60N mass (please refer to the image) we get $w_1-T={w_1\over g} \times a_1-----(1)$ From the free body diagram of 30N mass (please refer to the image) we get $T-w_2 ={w_2\over g} \times a_2-----(2)$ Adding (1) and (2) we get $w_1-T+ T-w_2= {w_1\over g} \times a_1+ {w_2\over g }\times a_2$ $60-30=({60\over9.81}+{30\over9.81})a_1$ $a_1={9.81\over3}=3.27m/s^2$ We know from kinematic relations $v^2-u^2=2as$ Here initial velocity $u=0m/s$ Acceleration $a_1=3.27m/s^2$ Displacement $s=h=2m$ $v_1^2=2\times3.27\times2=13.08m/s$ $v_1=3.62m/s$ So the speed at which $60N$ mass will hit the ground is $3.62m/s$