Verified answer

Use conservation of energy:

gravitational potential energy(GPE) lost = kinetic energy (KE) gained

GPE lost by m₂ = KE gained by m₁ +KE gained by m₂ + KE gained by pulley.

m₂gh = ½m₁v² + ½m₂v² + ½Iω²

Since ω = vR: m₂gh = ½m₁v² + ½m₂v² + ½I(v²/R²)

2m₂gh = v²(m₁ + m₂ + I/R²)

v = √[2m₂gh/((m₁ + m₂ + I/R²)]

Physics Torque..

First, draw the body mass diagram of each object.

Using Newton's Law

for object 1 : sigma_F = m*a --> T1 = m1 * a (eq 1)

for object 2 : sigma_F = m*a --> T2 = W2 - m2 * a (equation 2)

sigma_torque = moment_of_inertia x (linear_acceleration / R)

(R x T2) - (R x T1) = 2/5 M*R^2 x (a/R) (equation 3)--> T2 = tension of m2, T1 = tension of m1, M = mass of pulley

substituting eq 1 and 2 to eq 3, we got :

R*(T2-T1)=2/5 M*R^2 x (a/R)

R (W-m2*a-m1*a)=2/5 M*R^2 x (a/R)

eliminating R :

(W-m2*a-m1*a)=2/5 M*R x (a)

thus, a equal

a =(2/5 M + m2 - m1) / (m2 * g)

speed after moving h distance :

V_end^2 = V_start^2 + 2 * a * h

V_end^2 = 0+ 2*(2/5 M + m2 - m1) / (m2 * g) *h

V_end^2 = 2*h*(2/5 M + m2 - m1) / (m2 * g)

v_end = sqrt(2*h*(2/5 M + m2 - m1) / (m2 * g))