Question

2. Calculate the terminal speed for (a) a raindrop (radius =0.1mm) and (3%) (b ) a gold ball (radius = 35 mm, weight= 1950 g) (3%) (c) if a raindrop is released from the sky (h=10000 m) at t=0 s, then a golden ball is released from the some position at t =3 S, Will golden ball touch the raindrop in the sky? (4%) hint: assume drag coefficient = 0.5 t=0 t=3 te?

Answer 1

a fluid experiences (a) Any object moving through drag. that can be expressed as ed → dorag coefficient pas density of fluid velocity A characteristic frontal Fd= cd p v²A 2 vy area we can write the drop equation of motion the rain mdo mg - Fd dt do dt V Cd P v za ng - 2 when terminal velocity has been reached there will be no change in velocity with time or do at

cd P A 2. mg- 2 - 2 mg u APA here P = H = Tr2 = density of air = 1.2 kg/m3 3,14x (0.1 x 1587² m² 4 hrs Iwater * Pwater = 10 kg/m3 and - g m/sec² cd= VE = 2x yar x 9.8 x103 r= 0.1 x 10 m 0.5x 1.2 XT r2 2x4 X 0.1X103 x9,8X103 0.583 X 1.2 Vt = - 2.086 m/sec

(B)

() we can use the same formula in equation to &t= 2 mg Cd PA 1950g = 1.95 kg J = 9,8 Cd = 0.5 Pa 1.2 kg/m² A = πr? r= 35810 m y Vt 2x1.9589,8 0,5X 1.2x 3.1ux (35X16*)? > Ut 128.65 m/sec our equation of motion in en © is mdo dt CAPA 0² mg 2 do at Cd PA g - 12 2mg

do lee do g dt dt vt do - v2 VE At sino do= Acoso do ut it coso do costo sort ja secondo sot T vt ln (seco + tano) gtto /t ut la sino elgt + 1 - coso 1- 192 = gt tc ve ut ln utt v 11 gttc 12

5 . ln (Vetv utv gt + c for rain drop at t=0 v=0 8.0 to Vtr Verin (vest) = at tr 28th wth Y Utut v = (vt v) Te 28t/otr, 1) 28/0th + 1) st/ver gtatr v- & tn ghar + v tank (141) tr
time taken by the raindrop to reach terminal velocity can be approximated as

t = 3xV_tr/g                    since tanh(3) = 0.995, therefore at t = 3V_tr/g , v is approximately equal to V_tr

= 0.638 sec

almost immediately raindrop will start falling with its terminal velocity.

and its total flight time will be approximately

T_r = 10000/2.086

$\Rightarrow$T_r = 4793 seconds

at t= 3, for golden ball using equation & v- 0 Utoln uto g. 3 tc Ito > Сс - 3 g Uto In vittu Utov g (t-3) = ) V V to tanh g (t-3) Ito

similarly, time taken by the golden ball to reach terminal velocity can also be approximated by

t-3 = 3xV_tb/g

t = 3+3x128.65/9.8

t = 42.38 sec

now the ball has traveled a certain distance while achieving the terminal velocity lets say it is h

now it has to travel 10000-h distance with terminal velocity.

Therefore,

Total flight time = time taken to reach terminal velocity + time taken to travel remaining 10000-h meters with velocity v_tb

T_b = 42.38+(10000-h)/128.65

$\Rightarrow$ T_b = 42.38 + 77.73 - h/128.65

$\Rightarrow$ T_b = 120.11 - h/128.65

that is less than 120.11 seconds.

clearly the golden ball takes less time to reach the ground than the raindrop even though it started its flight 3-seconds late, therefore the golden ball has to touch the raindrop in the sky.