Small Raindrop, Big Raindrop

As the particle falls down with a non-constant mass a thrust force acts upwards opposite to the motion of magnitude $F_t = \dfrac{dm}{dt}v$ .

Since we have $\displaystyle \dfrac{dm}{dt}=5\times 10^{-3}$ , on integration it gives $m(t)=(5t+10)\times 10^{-3}$ .

Now we have the equation of motion of the particle as :

$\displaystyle \begin{aligned} &a(t) = g-\dfrac{1}{m}\dfrac{dm}{dt}v(t) \\ & v'(t) = g-\dfrac{5v(t)}{5t+10} \\ & v'(t)(5t+10)+5v(t)=g(5t+10) \\ & (5t+10)v(t)=\int_0^t g(5t+10)dt \\ & v(t)=\dfrac{5t(t+4)}{t+2}\end{aligned}$

Putting $t=4$ we get velocity as $\displaystyle \boxed{\dfrac{80}{3}}\approx 26.67$

$p_{f}-p_{i}=\displaystyle \int_{t_i}^{t_f} {F(t)dt}$

Where:

$g=10$ $m/s^2$

$F(t)=m(t)g=10(0.01+0.005t)$ $kg$

$t_i =0$ , $t_f=4$ $s$

$m_{f}=0.01+0.005t_{f}=0.01+0.005 \times 4=0.003$ $kg$

$p_{i}=0$ (since initially the raindrop is at rest), $p_{f}=m_{f} v=0.03v$

Then,

$\begin{aligned} 0.03v-0&=\displaystyle \int_{0}^{4} 10(0.01+0.005t) \space dt \\ 0.03v&=10(0.01t+0.0025t^2)|_{0}^{4} \\ 0.03v&=0.8 \\ v&=\boxed{26.67} \space m/s^2 \end{aligned}$