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Math Help - differential equation,seperation of variables solve for t(vertical projection)help

  1. #1
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    differential equation,seperation of variables solve for t(vertical projection)help

    example a stone is dropped from a bridge against quadratic air resistance and im deriving an expression of the stones velocity in terms of its position.
    mv dv/dx=mg-mg-kv^2
    integral v/g-kv^2=integral 1 dx k=6.2x10^-3
    -1/2kln(g-kv^2)=x+A
    g-kv^2=e^(-2kx-2kA)=Be^-2kx
    g-kv^2=ge^-2kx
    v=sqroot g/k sqroot 1-e^-2kx (distance stone falls is 77m)
    plugging every thing in
    v=sqroot 9.81/6.2x10^-3 sqroot 1-e^-2x6.2x10^-3x77=31.2 ms

    now how do i solve a similar problem, a ball is projected vertically upwards with a speed of 20ms and it reaches a max height of 17.37m and i need to find in the method i have shown the time it took to reach its max height and k=0.009

    if some one can show me how its done thanks ,but i want to solve it in the same kind of method ive shown no partial fractions,i have actually been told its not that hard to solve in that way.any help thanks
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  2. #2
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    Quote Originally Posted by puddy View Post
    example a stone is dropped from a bridge against quadratic air resistance and im deriving an expression of the stones velocity in terms of its position.
    mv dv/dx=mg-mg-kv^2
    integral v/g-kv^2=integral 1 dx k=6.2x10^-3
    -1/2kln(g-kv^2)=x+A
    g-kv^2=e^(-2kx-2kA)=Be^-2kx
    g-kv^2=ge^-2kx
    v=sqroot g/k sqroot 1-e^-2kx (distance stone falls is 77m)
    plugging every thing in
    v=sqroot 9.81/6.2x10^-3 sqroot 1-e^-2x6.2x10^-3x77=31.2 ms

    now how do i solve a similar problem, a ball is projected vertically upwards with a speed of 20ms and it reaches a max height of 17.37m and i need to find in the method i have shown the time it took to reach its max height and k=0.009

    if some one can show me how its done thanks ,but i want to solve it in the same kind of method ive shown no partial fractions,i have actually been told its not that hard to solve in that way.any help thanks
    mv dv/dx= -mg-kv^2. See the difference? Understand why? Now integrate etc.
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    help again

    well can you solve it for me thats the reason why i came on here, show me how its done ?
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    Quote Originally Posted by puddy View Post
    well can you solve it for me thats the reason why i came on here, show me how its done ?
    After re-arranging the differential equation, you have

    \frac{dv}{dx} = - \frac{(mg + kv^2)}{mv} \Rightarrow \frac{dx}{dv} = \frac{-mv}{mg + kv^2}, subject to the boundary condition x = 0, v = 0.

    Now integrate etc.

    (When studying differential equations it is assumed that you know how to integrate).
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