Question

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?

Answer 1

see below

Explanation:

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

`F = mg - (1 + v/10) = m dot v`

As `m = 0.2`:

`dot v = (2g - 10 - v)/2 = (alpha - v)/2`

This separates:

`(dv)/(alpha - v) = 1/2 dt`

And integrates as:

`- ln (alpha - v ) = 1/2t + C, qquad alpha >0, qquad 0 le v lt alpha`

`v(0) = 0`

`- ln(alpha ) = C`

`ln (alpha/(alpha - v)) = 1/2t`

`alpha/(alpha - v) = e^(1/2t)`

`v = alpha (1 - e^(-1/2t))`

Terminal velocity

`lim_(t to oo) v = lim_(t to oo) alpha (1 - e^(-1/2t))`

`= alpha = 2g - 10 = 9.6 "m/s"`

Time `tau` to reach half of terminal velocity

`1/2 = (1 - e^(-1/2tau))`

`-1/2tau = ln(1/2)`

`tau = 2 ln( 2) " sec"`

Answer 2

The terminal velocity is `=9.6ms^-1`. The time to reach half the speed is `=1.39s`

Explanation:

The mass of the balloon is `m=0.2kg`

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

`mg-R=m(dv)/dt`

`mg-(1+0.1v)=m(dv)/dt`

The acceleration due to gravity is `g=9.8ms^-2`

When the terminal velocity is reached, the velocity is constant

Therefore,

`(dv)/dt=0`

`mg-(1+0.1v)=0`

`1+0.1v=0.2*9.8=1.96`

`0.1v=1.96-1=0.96`

`v=0.96/0.1=9.6ms^-1`

Half the terminal velocity is `v_1=9.6/2=4.8ms^-1`

The initial velocity is `u=0ms^-1`

Solve the differential equation

`mg-(1+0.1v)=m(dv)/dt`

`(dv)/(mg-1-0.1v)=1/mdt`

`int_0^t1/mdt=int_0^4.8(dv)/(mg-1-0.1v)`

`1/mt=[-10ln(mg-1-0.1v)]_0^4.8`

`5t=[-10ln(0.96-0.1v)]_0^4.8`

`=-10ln(0.48)+10ln(0.96)=10*0.69=6.93`

The time is `=6.93/5=1.39s`