A hot air balloon of mass $M$ is equipped with a bag of sand of mass $m=m(t)$ which decreases in time as the sand is gradually released. In addition to gravity the balloon experiences a constant upwards buoyancy force $T$ and we neglect air resistance effects. Show that if $v(t)$ is the upward speed of the balloon then

$$(M+m) \frac{d v}{d t}=T-(M+m) g .$$

Initially at $t=0$ the mass of sand is $m(0)=m_{0}$ and the balloon is at rest in equilibrium. Subsequently the sand is released at a constant rate and is depleted in a time $t_{0}$. Show that the speed of the balloon at time $t_{0}$ is

$$g t_{0}\left(\left(1+\frac{M}{m_{0}}\right) \ln \left(1+\frac{m_{0}}{M}\right)-1\right)$$

[You may use without proof the indefinite integral $\int t /(A-t) d t=-t-A \ln |A-t|+C .$ ]