walt asked in Science & MathematicsPhysics · 1 decade ago

Motion of a projectile with air resistance?

Please help.

A projectile is thrown verticali into the air with initial velocity of 20m per second (v0) at t=0

Using the quadratic model for air resistance and using a=dv/dt find t in terms of v, b, g and v0 and hence find the time taken for the projectile to reach the maximum hight.

b^2 = mg/0.2D^2

m = 0.02kg, D = 0.03m, g = 9.81ms^-2, vo= 20ms^-1

In vector form, with origin at point of projection, I use

ma = -mg-(0.2D^2)(V^2)

mdv/dt = -mg-(0.2D^2)(V^2)

m/-mg-(0.2D^2)(V^2) dv = 1 dt

1/-g-k(V^2) dv = 1 dt , k=-(0.2D^2)/m

I intergrate and get

-1/2k(In(-g-k(V^2)) = t + A

Am I correct ?

If so wy cant I get the correct answer ? it sould be around 2 seconds.

Also using a = vdv/dx how do I find the hight.It should be 17.37m

Many thanks

1 Answer

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  • 1 decade ago
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    sorry, its not correct

    You made an error in integration.

    Using your notation:

    dv/dt = - g - k · v²

    <=>

    ∫ 1/(g+k·v²) dv = - ∫ dt

    Integration on LHS leads to the inverse tangent: ∫1/(1+z²) dz = arctan(z)

    =>

    ∫ 1/(1+(√(k/g)·v)²) dv = - g·∫ dt

    <=>

    √(g/k)·arctan(√(k/g)·v) = - g·t + c

    <=>

    arctan(√(k/g)·v) = C - √(g·k)·t

    where C=√(k/g)·C

    Apply initial condition to evaluate C

    v(t=0) = v₀

    <=>

    arctan(√(k/g)·v₀) = C - √(g·k)·0

    <=>

    arctan(√(k/g)·v₀) = C

    =>

    arctan(√(k/g)·v) = arctan(√(k/g)·v₀) - √(g·k)·t

    <=>

    v = √(g/k)·tan[arctan(√(k/g)·v₀) - √(g·k)·t ]

    At maximum height velocity becomes zero:

    v(t_max) = 0

    <=>

    arctan(√(k/g)·0) = arctan(√(k/g)·v₀) - √(g·k)·t_max

    <=>

    t_max = arctan(√(k/g)·v₀) / √(g·k)

    I would calculate it, but I'm a bit puzzled from your formula for the air resistance, particularly the units.

    I doubt that it is correct. Whatever is correct, k must have dimension m^-1 to be consistent with the units of v and g.

    For second part

    a = dv/dx

    v dv/dx = - g - k·v²

    <=>

    ∫ v/(g+k·v²) dv = - ∫ dx

    <=>

    (1/2)·∫ 2·v/(g+k·v²) dv = - ∫ dx

    (1/(2·k))·ln(g+k·v²) = -x + d

    <=>

    ln(g+k·v²) = 2·k·d - 2·k·x

    <=>

    g+k·v² = D·e^(-2·k·x)

    where D = e^(2·k·d)

    =>

    Apply initial condition to evaluate D

    v(x=0) = v₀

    <=>

    g+k·v₀² = D·e^(-k·0)

    <=>

    D = (g+k·v₀²)

    =>

    g+k·v² = (g+k·v₀²)·e^(-2·k·x)

    <=>

    v = √[(g+k·v₀²)·e^(-2·k·x) - g]

    At maximum height velocity equals zero:

    v(x_max) = 0

    <=>

    √[(g+k·v₀²)·e^(-2·k·x_max) - g] = 0

    <=>

    e^(-2·k·x_max) = g/(g+k·v₀²)

    <=>

    e^(2·k·x_max) = (g+k·v₀²)/g

    <=>

    x_max = ln[1+(k/g)·v₀²] /(2·k)

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