How Long is the Stone in the Air?Date: 8/5/96 at 22:29:45 From: Anonymous Subject: Time Of Stone's Flight A stone is thrown from the ground at u m/s and returns to the ground at v m/s. Assuming air resistance is proportional to the square of the velocity, the constant of proportionality being k, what is the total time of flight? (in terms of g, v, u, and k) Date: 8/7/96 at 7:55:7 From: Doctor Anthony Subject: Re: Time Of Stone's Flight The equation of motion on upward flight is: (Here I am using v as the variable velocity. I will use U and V as launch and return velocities) m*dv/dt = -mg - mkv^2 INT[dv/(g+kv^2)] = -INT[dt] (1/k)INT[dv/(g/k + v^2)] = -INT[dt] (1/k)sqrt(k/g)tan^(-1){v/sqrt(g/k)} = -t + const sqrt(1/(kg))tan^(-1){sqrt(k/g)*v} = -t + const At t=0 v=U so const = sqrt(1/(kg))tan^(-1){sqrt(k/g)*U} Now at the top of the flight path v=0 and so the time to reach the top is given by: t1 = sqrt(1/(kg))tan^(-1){sqrt(k/g)*U} For the downward journey the equation of motion is: m*dv/dt = mg - mkv^2 INT[dv/(g-kv^2)] = INT[dt] (1/k)INT[dv/(g/k - v^2)] = INT[dt] (1/k)sqrt(k/g)tanh^(-1){v/sqrt(g/k)} = t + const At t=0, v=0 so the constant = 0 And the time to reach speed V will be given by: t2 = sqrt(1/(kg))tanh^(-1){sqrt(k/g)*V} And so total time of flight is given by t1 + t2. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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