"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)(adsbygoogle = window.adsbygoogle || []).push({});

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

[tex]\frac{dx}{dt}=v=0[/tex]

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]

gives

[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.

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# First-order differential equation

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