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Mechanics: Vertical projection with air resistance

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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.

https://brainmass.com/physics/resistance/mechanics-vertical-projection-air-resistance-201873

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) ∫[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) ∫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) ∫[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) ∫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)
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Or v = v0vt/√(vt2+v02)

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