# A meteorological balloon of mass 0.2 Kg falls vertically from rest, the total resistance at V m/s is approximately (1+0.1V)N taking g=9.8 m/s^2# . Calculate: a) The terminal velocity? b) Time taken for the balloon to reach half of the terminal velocity?

Jun 8, 2018

see below

#### Explanation:

For motion, with positive direction vertically downwards, Newton's 2nd Law:

$F = m g - \left(1 + \frac{v}{10}\right) = m \dot{v}$

As $m = 0.2$:

$\dot{v} = \frac{2 g - 10 - v}{2} = \frac{\alpha - v}{2}$

This separates:

$\frac{\mathrm{dv}}{\alpha - v} = \frac{1}{2} \mathrm{dt}$

And integrates as:

$- \ln \left(\alpha - v\right) = \frac{1}{2} t + C , q \quad \alpha > 0 , q \quad 0 \le v < \alpha$

$v \left(0\right) = 0$

$- \ln \left(\alpha\right) = C$

$\ln \left(\frac{\alpha}{\alpha - v}\right) = \frac{1}{2} t$

$\frac{\alpha}{\alpha - v} = {e}^{\frac{1}{2} t}$

$v = \alpha \left(1 - {e}^{- \frac{1}{2} t}\right)$

Terminal velocity

${\lim}_{t \to \infty} v = {\lim}_{t \to \infty} \alpha \left(1 - {e}^{- \frac{1}{2} t}\right)$

$= \alpha = 2 g - 10 = 9.6 \text{m/s}$

Time $\tau$ to reach half of terminal velocity

$\frac{1}{2} = \left(1 - {e}^{- \frac{1}{2} \tau}\right)$

$- \frac{1}{2} \tau = \ln \left(\frac{1}{2}\right)$

$\tau = 2 \ln \left(2\right) \text{ sec}$

Jun 8, 2018

The terminal velocity is $= 9.6 m {s}^{-} 1$. The time to reach half the speed is $= 1.39 s$

#### Explanation:

The mass of the balloon is $m = 0.2 k g$

The net force on the balloon according to Newton's second law is

$m g - R = m \frac{\mathrm{dv}}{\mathrm{dt}}$

$m g - \left(1 + 0.1 v\right) = m \frac{\mathrm{dv}}{\mathrm{dt}}$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

When the terminal velocity is reached, the velocity is constant

Therefore,

$\frac{\mathrm{dv}}{\mathrm{dt}} = 0$

$m g - \left(1 + 0.1 v\right) = 0$

$1 + 0.1 v = 0.2 \cdot 9.8 = 1.96$

$0.1 v = 1.96 - 1 = 0.96$

$v = \frac{0.96}{0.1} = 9.6 m {s}^{-} 1$

Half the terminal velocity is ${v}_{1} = \frac{9.6}{2} = 4.8 m {s}^{-} 1$

The initial velocity is $u = 0 m {s}^{-} 1$

Solve the differential equation

$m g - \left(1 + 0.1 v\right) = m \frac{\mathrm{dv}}{\mathrm{dt}}$

$\frac{\mathrm{dv}}{m g - 1 - 0.1 v} = \frac{1}{m} \mathrm{dt}$

${\int}_{0}^{t} \frac{1}{m} \mathrm{dt} = {\int}_{0}^{4.8} \frac{\mathrm{dv}}{m g - 1 - 0.1 v}$

$\frac{1}{m} t = {\left[- 10 \ln \left(m g - 1 - 0.1 v\right)\right]}_{0}^{4.8}$

$5 t = {\left[- 10 \ln \left(0.96 - 0.1 v\right)\right]}_{0}^{4.8}$

$= - 10 \ln \left(0.48\right) + 10 \ln \left(0.96\right) = 10 \cdot 0.69 = 6.93$

The time is $= \frac{6.93}{5} = 1.39 s$