# Vertical Trajectory Calculation for Quadratic Drag of form -cv2

Air drag will depend upon the area of the object in motion and the resulting trajectory will depend upon its mass. For that reason the sphere calculation is typically the starting point for modeling motion with quadratic drag. Departures from spherical geometry can then be tried by varying the drag coefficient.

But a trajectory can be modeled with just the general form -cv2 for the quadratic drag. This is an example of a vertical trajectory calculation.

Assume the object has a drag coefficient c = Ns2/m2 and a mass of m = gm.

Under these conditions, it's terminal velocity will be
vt = m/s
vt = km/hr
vt = mi/hr

and its characteristic time is
τ = s.

If this sphere is launched vertically with a velocity of
v0 = m/s = ft/s

then on the way up at height y1 = m = ft
it will have velocity v1 = m/s = ft/s

It will reach a peak height ypeak = m = ft
at time tpeak = s

On the way down at height y2 = m = ft
it will have velocity v2 = m/s = ft/s

It will reach the ground at velocity
vimpact = m/s = ft/s = km/hr = mi/hr
at time timpact = s.

For comparison, if there were no air friction the projectile would have reached height
h = m = ft
and would impact with the original launch velocity at time t = s.

Notes: This is not as practical a calculation form as the calculation for a sphere in air, but could be useful for obtaining an effective drag coefficient. If we were to model the trajectory of a 100 gram spherical rock of radius 2cm (density about 3000 kg/m3 or specific gravity 3) through air with density 1.29 kg/m3, that alternate form of the calculation would lead to a drag coefficient c = .0004 Ns2/m2. If you leave the values of c and radius at zero, they will default to the values for this rock. That will give you an example which you could then compare to the sphere calculation.

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Fluid friction

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