Rhea asked in Science & MathematicsPhysics · 9 months ago

# Terminal Velocity Help!?

Terminal velocity is the maximum speed that a body falling through air can reach due to air resistance. Terminal velocity is given by the formula v Subscript t Baseline equals StartRoot StartFraction 2 mg Over Upper C rho Upper A EndFraction EndRoot

​, where m is the mass of the falling​ object, g is acceleration due to gravity ​(almost equals9.81 meters per second squared​), C is a drag coefficient with 0.5 less than or equals Upper C less than or equals 1.0​, rho is the density of air ​(almost equals1.2 ​kg/m cubed​), and A is the​ cross-sectional area of the object. Suppose that a raindrop falls from the sky. The mass of the raindrop is given by m equals four thirds pi r cubed d

where r is its radius and d equals 1000 kg divided by m cubed. The​ cross-sectional area of the raindrop is Upper A equals pi r squared.

Relevance
• Anonymous
9 months ago

You have the equation, what's the issue?

• Vaman
Lv 7
9 months ago

I can not get this.

The terminal velocity

v= c/g. Make v and c as the function of x. Take the average velocity. If you add the area, then it will be

v= c/(gx)

It is a tricky question. We need a formaula for c.If you put c= ax+b . at x=0, c=0.5, at x=1, ground level c=1, This gives

c= 0.5 x +0.5. May be you can try

dv= (dx)/(2gx) , Integrate between the limit x=0 a and 1 and see this tallies with the answer. Change the limits to 1 and 2 and adjust a and b and integrate. This removes the singularity that comes at x=0. Try your luck.

• 9 months ago

vt( at C = 1) = sqrt( 2mg/( C rho A) )

= sqrt( 2 * 4/3 pi() r^3 * 1000 * 9.81/( 1 * 1.2 * pi() r^2) )

= sqrt(8/3 r * 1000 * 9.81/ 1.2) =sqrt( 21800 r)

-> sqrt( 21800 r) < vt < sqrt( 2 * 21800 r)

• 9 months ago

• oubaas
Lv 7
9 months ago

Vt = √2*m*g/(C*ρa*A)

mass m = Vol*ρw = 0.52360*d^3*1000

C = 0.75

ρa = 1.25

A = 0.7854d^2

Vt = √2*9*80665*523.6*d^3/(0.75*1.25*0.7854*d^2 = 118.10√d m/sec