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First-order differential equation

  1. Jun 9, 2005 #1
    "A ball with mass 0.15 kg is thrown upward with initial velocity 20 m/s from the roof of a building 30 m heigh. There is a force due to air resistance of [tex]\left| v \right|/30[/tex] where the velocity [tex]v[/tex] is measured in m/s. Find the time that the ball hits the ground." (Answer: 5.129 s)

    I split the problem into two parts---namely, the way up and down. From the former, I get the time for the ball to reach the maximum height. Then, I use the latter to find the answer.

    If we measure [tex]x[/tex] positively upward from the ground, then

    [tex]m\frac{dv}{dt}=-mg-\frac{v}{30}, \qquad v(0) = v_0, \qquad x(0)=x_0[/tex]

    [tex]\frac{dv}{dt} + \frac{v}{30m}=-g[/tex]

    The Method of Integrating Factors gives

    [tex]\mu = \exp \left( \frac{1}{30m} \int dt \right) =\exp \left( \frac{t}{30m} \right)[/tex]

    [tex]v(t)=\exp \left( -\frac{t}{30m} \right) \int -g\exp \left( \frac{t}{30m} \right) \: dt[/tex]

    [tex]v(t)=\exp \left( -\frac{t}{30m} \right) \left[ -g(30m) \exp \left( \frac{t}{30m} \right) + \mathrm{C} \right][/tex]

    [tex]v(t)=-30gm + \mathrm{C} \exp \left( -\frac{t}{30m} \right)[/tex]

    Next, we apply the initial condition in order to find the constant.

    [tex]v(0) = v_0 \Rightarrow \mathrm{C} = v_0 + 30 gm[/tex]

    Thus, we obtain

    [tex]v(t) = -30gm + \left( v_0 + 30 gm \right) \exp \left( -\frac{t}{30m} \right)[/tex]

    At the maximum height, we have


    which gives

    [tex]t_{\mbox{up}} = 30m \ln \left( \frac{v_0}{30gm} + 1 \right)[/tex]

    The way down is described as follows:

    [tex]m\frac{dv}{dt}=-mg+\frac{v}{30}, \qquad v\left( t_{\mbox{up}} \right) = 0, \qquad x\left( t_{\mbox{up}} \right) = x_{\mbox{max}}[/tex]

    [tex]\frac{dv}{dt} - \frac{v}{30m}=-g[/tex]

    The Method of Integrating Factors gives

    [tex]\mu = \exp \left( -\frac{1}{30m} \int dt \right) =\exp \left( -\frac{t}{30m} \right)[/tex]

    [tex]v(t)=\exp \left( \frac{t}{30m} \right) \int -g \exp \left( -\frac{t}{30m} \right) \: dt[/tex]

    [tex]v(t)=\exp \left( \frac{t}{30m} \right) \left[ -g (-30m) \exp \left( -\frac{t}{30m} \right) + \mathrm{C} \right][/tex]

    [tex]v(t)= 30mg + \mathrm{C}\exp \left( \frac{t}{30m} \right)[/tex]

    Next, we apply the initial condition in order to find the constant.

    [tex]v\left( t_{\mbox{up}} \right) = 0 \Rightarrow \mathrm{C} = -\frac{(30mg) ^2}{v_0 + 30mg}[/tex]

    Thus, we obtain

    [tex]v(t)= 30mg -\frac{(30mg) ^2}{v_0 + 30mg} \exp \left( \frac{t}{30m} \right)[/tex]

    and so

    [tex]\int v(t) \: dt = 30mg\int dt -\frac{(30mg) ^2}{v_0 + 30mg} \int \exp \left( \frac{t}{30m} \right) \: dt[/tex]

    [tex]x(t)= 30mgt -\frac{(30m) ^3 g^2}{v_0 + 30mg} \exp \left( \frac{t}{30m} \right) + \mathrm{C}[/tex]

    Next, we apply the initial condition in order to find the constant.

    [tex]x\left( t_{\mbox{up}} \right) = x_{\mbox{max}} \Rightarrow \mathrm{C} = x_{\mbox{max}} + (30m)^2 g - (30m)^2 g \ln \left( \frac{v_0}{30mg} + 1 \right)[/tex]

    Thus, we obtain

    [tex]x(t)= 30mgt -\frac{(30m) ^3 g^2}{v_0 + 30mg} \exp \left( \frac{t}{30m} \right) + x_{\mbox{max}} + (30m)^2 g - (30m)^2 g \ln \left( \frac{v_0}{30mg} + 1 \right)[/tex]

    When the ball hits the ground, [tex]x(t)=0[/tex]. Solving for [tex]t[/tex] gives [tex]t \approx 4.429 \mbox{ s}[/tex]. The result is clearly wrong, but there is something even more weird than that. Using [tex]x(t)[/tex] (way up) gives the right answer! In other words

    [tex]v(t) = -30gm + \left( v_0 + 30 gm \right) \exp \left( -\frac{t}{30m} \right)[/tex]


    [tex]\int v(t) \: dt = -30gm \int dt + \left( v_0 + 30 gm \right) \int \exp \left( -\frac{t}{30m} \right) \: dt[/tex]

    [tex]x(t) = -30gmt -30m \left( v_0 + 30 gm \right) \exp \left( -\frac{t}{30m} \right) + \mathrm{C}[/tex]

    Next, we apply the initial condition in order to find the constant.

    [tex]x(0) = x_0 \Rightarrow \mathrm{C} = x_0 + 30m \left( v_0 + 30gm \right)[/tex]

    [tex]x(t) = x_0 -30gmt + 30m \left( v_0 + 30 gm \right) \left[ 1 - \exp \left( -\frac{t}{30m} \right) \right][/tex]

    Again, setting [tex]x(t)=0[/tex] and solving for [tex]t[/tex] gives [tex]t \approx 5.129 \mbox{ s}[/tex]. Can anybody please explain me why this is happening? From what I have, it seems that the differential equation that models the ball falling is not necessary, which is strange.

    Any help is highly appreciated.
    Last edited: Jun 9, 2005
  2. jcsd
  3. Jun 9, 2005 #2


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    Science Advisor
    Homework Helper

    It's good that you noted the resistance is always opposite to the direction of motion.
    You actually made it too difficult for yourself. You need only one equation

    which is your first one.
    The minus sign in front of the resistance term automatically makes it opposed to the direction of motion.
    Since you've chosen the positive direction upwards. Now, when the ball rises v is positive so the resistance force is negative (downwards).
    When the ball falls v is negative and the resistance force is positive (upwards). So you don't need to fuss with signs. You probably mistakenly assumed v to be always positive or something.
  4. Jun 9, 2005 #3
    Oh, sure! The reason why I posted this question is because the other day I was solving a similar DE where there was a [tex]v^2[/tex] term. It's clear to me now.

  5. Sep 6, 2011 #4
    Hello, I have just worked this problem and had issue with getting 5.129s as an answer also. The only thing i don't understand is why the differential equation has -mg and not mg. Shouldn't this term be mg if the ball is falling because gravity is speeding it up not slowing it down?

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