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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
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Associated Topics:
High School Physics/Chemistry

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