Solve a and b, please. 1. A particle is projected vertically upward in a constan

Question

Solve a and b, please.

1. A particle is projected vertically upward in a constant gravitational field with an initial speed vo. There is a quadratic drag force of magnitude co2 acting on the particle. (a) (4 points) Show that the speed of the particle when it returns to the initial position is given by vovt } where ve is the terminal speed. (b) (2 points) Show that as the drag coefficient approaches zero your answer in part (a) reduces to the well known freshman physics result vj -vo Hint: Use a Taylor series expansion (what is your "small parameter"?)

Explanation / Answer

1. given initial sped = vo

drag = -cv^2

so while going upm time taken = t1

while going down, time taken = t2

for going up

mg + cv^2 + mv' = 0

m*dv/dt = -(cv^2 + mg)

m*dv/(v^2 + mg/c) = -cdt

integrating

m*arctan(v/sqrt(mg/c))/sqrt(mg/c) + k = -ct

at t= 0

k = -m*arctan(vo/sqrt(mg/c))/sqrt(mg/c)

hence

-ct = m*arctan(v/sqrt(mg/c))/sqrt(mg/c)-m*arctan(vo/sqrt(mg/c))/sqrt(mg/c)

-ct = m(arctan(v/sqrt(mg/c)) - arctan(vo/sqrt(mg/c)))/sqrt(mg/c) -- (1)

now, when v = 0

t = sqrt(m/gc)(arctan(vo/sqrt(mg/c)))

for coming down

cv^2 - mg + mv' = 0

mdv/dt = (mg - cv^2)

mdv/(mg/c - v^2) = cdt

integrating

m*arctanh(v/sqrt(mg/c))/sqrt(mg/c) + k = ct

now at t = sqrt(m/gc)(arctan(vo/sqrt(mg/c))), v = 0

hence

k = c*sqrt(m/gc)(arctan(vo/sqrt(mg/c)))

so,

m*arctanh(v/sqrt(mg/c))/sqrt(mg/c) + sqrt(mc/g)(arctan(vo/sqrt(mg/c))) = ct

dv = sqrt(mg/c)tanh[sqrt(gc/m)t - (arctan(vo/sqrt(mg/c))) ]*dt -- (2)

from 1 and 2 we find the amximum height, equate those and finally get the expression

vf = vo*vt/sqrt(vo^2 + vt^2)

where vt is terminal speed

b. when c = 0

then

vf = vo*vt/vt(1 + (vo/vt)^2)

from taylor series

vf = vo(1 + vo/2vt)

now, vt = infinite, when c = 0

hence

vf = vo

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