# Mechanics: Vertical projection with air resistance

A particle is projected vertically upward in a constant gravitational field with an initial speed of v_0. Show that if there is a retarding force proportional to the square of the instantaneous speed, the speed of the particle when it returns to the initial position is:

(v_0*v_t)/[(v_0)^2 + (v_t)^2]^1/2

where v_t is the terminal speed.

Please show the frame of reference used and all steps.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please refer to the attachment for complete solution.

As the particle travels upwards, it is subjected to downward forces weight mg and air drag fd where fd = kv2 where v is the instantaneous speed and k is the proportionality constant.

Net force on the particle F = mg + kv2

Or instantaneous acceleration a = (mg + kv2)/m .......(1)

Let the particle move up by a small distance dy in time dt and its velocity decreases from v to (v-dv)

Applying v2 - u2 = 2as to the particle for its displacement dy :

(v - dv)2 - v2 = 2[(mg + kv2)/m] dy

Or v2 + dv2 - 2v dv - v2 = 2[(mg + kv2)/m] dy

Ignoring dv2 as negligible we get : - v dv = [(mg + kv2)/m] dy

Or dy = - (m/k) [v/[(gm/k + v2)] dv

Or y = - (m/k) âˆ«[v/[(gm/k + v2)] dv + C where C is the constant of integration

Substituting (gm/k + v2) = x, dx = 2v dv, the above equation becomes :

y = - (m/k) âˆ«dx/2x + C

Or y = -(m/2k) logex + C

Or y = -(m/2k) loge(gm/k + v2) + C

As at y = 0, v = v0, 0 = -(m/2k)loge(gm/k + v02) + C or C = (m/2k)loge(gm/k + v02)

y = -(m/2k) loge(gm/k + v2) + (m/2k)loge(gm/k + v02)

At the highest point v = 0. Hence, displacement at the highest point is given by :

ymax = -(m/2k) loge(gm/k) + (m/2k)loge(gm/k + v02) = (m/2k)[loge(1+k/gm v02)] .....(1)

Downwards motion : For downward motion we consider the coordinate system as shown below :

0

fd

mg

ymax

As the particle travels downwards, the air drag acts upwards. Hence, net force on it becomes:

Net force on the particle F = mg - kv2

Instantaneous acceleration a = (mg - kv2)/m

Let the particle move down by a small distance dy in time dt and its velocity increases from v to (v+dv). Applying v2 - u2 = 2as we get :

(v + dv)2 - v2 = 2[(mg - kv2)/m]dy

Or v2 + dv2 + 2v dv - v2 = 2[(mg - kv2)/m] dy

Ignoring dv2 as negligible we get : v dv = [(mg - kv2)/m] dy

Or dy = (m/k) [v/[(gm/k - v2)] dv

Or y = (m/k) âˆ«[v/[(gm/k - v2)] dv + D where D is the constant of integration

Substituting (gm/k - v2) = x, dx = - 2v dv, the above equation becomes :

y = - (m/k) âˆ«dx/2x + D

Or y = -(m/2k) logex + D

Or y = -(m/2k) loge(gm/k - v2) + D

As at y = 0, v = 0, Or D = (m/2k)loge(gm/k)

y = -(m/2k) loge(gm/k - v2) + (m/2k)loge(gm/k)

To get the velocity at the origin (ground), we put y = ymax in the above :

ymax = -(m/2k) loge(gm/k - v2) + (m/2k)loge(gm/k)

Substituting for ymax from (1) :

(m/2k)[loge(1+k/gm v02)] = -(m/2k) loge(gm/k - v2) + (m/2k)loge(gm/k) ....(2)

At the terminal velocity net force on the particle is 0. Hence, mg = kvt2 or mg/k = vt2 and (2) becomes :

(m/2k)[loge(1+v02/vt2) = -(m/2k) loge(vt2 - v2) + (m/2k)loge(vt2)

loge(1+v02/vt2) = -loge(vt2 - v2) + loge(vt2)

Or (1+v02/vt2) = vt2/(vt2 - v2)

Or (vt2 - v2) = vt2/(1+v02/vt2) = vt4/(vt2+v02)

Or v2 = vt2 - vt4/(vt2+v02) = (vt4+v02vt2 - vt4)/(vt2+v02) = v02vt2/(vt2+v02)

_______

Or v = v0vt/âˆš(vt2+v02)

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