First-order differential equation for ball of mass

[DivGradCurl]

A ball with mass 0.15\mbox{ kg} is thrown upward with initial velocity 20\mbox{ m/s} from the roof of a building 30 \mbox{ m} high. There is a force due to air resistance of \frac{v^2}{1325}, where the velocity v is measured in \mbox{ m/s}.

I've deliberately disregarded the rest of the problem. My question is about the first-order differential equation I set up based upon the given information.

m\frac{dv}{dt}=-mg-\frac{v^2}{1325}, \qquad v(0)=v_0 \quad x(0)=x_0

\frac{dv}{dt} + \frac{v^2}{1325m} = -g

As far as I can see, the method of integrating factors does not work here. I don't know what to do. It is possible that I made a mistake in the D.E. set up.

Any help is highly appreciated.

 

[dextercioby]

HINT:Variables are separable.

Daniel.

 

[DivGradCurl]

Oh, I see. Thanks.

Here's what I have now:

\frac{dv}{dt}=-\frac{v^2}{1325m}-g

\frac{dv}{dt}=-\frac{\left( v^2 + 1325 mg \right)}{1325m}

\int \frac{dv}{v^2 + 1325 mg} = -\frac{1}{1325m}\int dt

On the left-hand side, the substitution v=\sqrt{1325 mg}\tan \theta gives

\int \frac{dv}{v^2 + 1325 mg} = \frac{\sqrt{1325mg}}{1325mg} \arctan \left( \frac{\sqrt{1325mg}}{1325mg}v \right) + \mathrm{C}

and so we have

\frac{\sqrt{1325mg}}{1325mg} \arctan \left( \frac{\sqrt{1325mg}}{1325mg}v \right) = -\frac{t}{1325m} + \mathrm{C}

v(0)=v_0 \Rightarrow \mathrm{C} = \frac{\sqrt{1325mg}}{1325mg} \arctan \left( \frac{\sqrt{1325mg}}{1325mg}v_0 \right)

 

[dextercioby]

Yes,it looks okay.

Daniel.

 

[arildno]

Keep in mind that this solution only holds on the way up...

 

[DivGradCurl]

I've got some more, but I'm not sure it is correct.

\frac{\sqrt{1325mg}}{1325mg} \arctan \left( \frac{\sqrt{1325mg}}{1325mg}v \right) = \frac{\sqrt{1325mg}}{1325mg} \arctan \left( \frac{\sqrt{1325mg}}{1325mg}v_0 \right) -\frac{t}{1325m}

\arctan \left( \frac{\sqrt{1325mg}}{1325mg}v \right) = \arctan \left( \frac{\sqrt{1325mg}}{1325mg}v_0 \right) -\frac{gt}{\sqrt{1325mg}}

v(t) = \sqrt{1325mg} \tan \left[ \arctan \left( \frac{\sqrt{1325mg}}{1325mg}v_0 \right) -\frac{gt}{\sqrt{1325mg}} \right]

x(t) = \sqrt{1325mg} \int \tan \left[ \arctan \left( \frac{\sqrt{1325mg}}{1325mg}v_0 \right) -\frac{gt}{\sqrt{1325mg}} \right] \: dt

Consider the integral

\int \tan (a-bt) \: dt

The substitution u=a-bt gives

\int \tan (a-bt) \: dt = - \frac{1}{b} \int \tan u \: du = - \frac{1}{b} \ln \left| \sec u \right| + \mathrm{C} = - \frac{1}{b} \ln \left| \sec (a-bt) \right| + \mathrm{C}

Thus, we have

x(t)= 1325m \ln \left| \sec \left[ \arctan \left( \frac{\sqrt{1325mg}}{1325mg} v_0 \right) -\frac{gt}{\sqrt{1325mg}} \right] \right| + \mathrm{C}

x(0)=x_0 \Rightarrow \mathrm{C} = x_0 + \frac{1325m}{2}\ln \left( \frac{1325mg}{1325mg + v_0 ^2} \right)

x(t)= 1325m \ln \left| \sec \left[ \arctan \left( \frac{\sqrt{1325mg}}{1325mg} v_0 \right) -\frac{gt}{\sqrt{1325mg}} \right] \right| + x_0 + \frac{1325m}{2}\ln \left( \frac{1325mg}{1325mg + v_0 ^2} \right)

Thanks

 

[arildno]

Remember that \cos(arctan(y))=\frac{1}{\sqrt{1+{y}^{2}}}

This will give you x(0)=x_{0} as it should..

 

[DivGradCurl]

That's what I did up there! Unfortunately, when I use the v(t) and x(t) above, I get some weird results.

Let's say I want to find the maximum height above the ground the ball reaches. That happens when \frac{dx}{dt}=v(t)=0. I simply need to solve for t and plug it into x(t). I get t \approx 1.9161467809752109 \mbox{ s} and x(t) \approx 11.438482854170548 \mbox{ m}. The answer in my textbook is x(t) \approx 48.562 \mbox{ m}.

Let's now suppose I want to find the time when the ball hits the ground. I set x(t)=0 and solve for t. The problem now is that I get complex numbers, while the answer in my textbook is t \approx 5.194 \mbox{ s}

There must be some mistake when I try to integrate v(t), but I don't know exactly where it is.

Thanks

 

[DivGradCurl]

First off, let me rectify my previous post. When I use the x(t) and v(t) above to find:

(a) the maximum height above the ground the ball reaches, I should have instead

v(t)=0. Then, solving for t gives

t\approx 1.9161467809752113 \mbox{ s}

and so

x(t) \approx 48.561517145829434 \mbox{ m}

(b) the time when the ball hits the ground, I should have instead

x(t)=0. Then, solving for t gives

t\approx 4.937745892290328 \mbox{ s}

The good news is that I get the exact same answers with the x(t) and v(t) I obtain with the aid of Mathematica's command DSolve, which solves differential equations. I just needed to enter this:

m\frac{dv}{dt}=-mg-\frac{v^2}{1325}, \qquad v(0)=v_0 \quad x(0)=x_0

which is the same I used to solve it myself. Thus, the mistake must either be in the differential equation set up I did or in the answers provided by the textbook. Let me reiterate what the textbook gives

(a) x(t) \approx 48.562 \mbox{ m}

(b) t \approx 5.194 \mbox{ s}

I still think it is much more likely that I made the mistake. Can anybody please help me figure this out?

Thanks

 

[arildno]

You got a) right then after all!

As for b), did you read my post 5?

Remember that air resistance is always in the direction OPPOSITE to the velocity the object has through the air!

Thus, the air resistance acting upon the object on the way down cannot be -kv^{2} where k is some constant gravity acts along the negative axis.

The correct quadratic air resitance law is: -k|v|v, where |v| is the speed of the object.
Thus, on its way down, Newton's second law reads in your case (remember v<0 now!):
m\frac{dv}{dt}=-mg+\frac{v^{2}}{1325}
Got it?

As a check, you'll end up with an expression involving the inverse of the hyperbolic tangent function, rather than arctan.

 

[DivGradCurl]

Now it makes sense! Thank you

m\frac{dv}{dt}=-mg+\frac{v^2}{1325}

\frac{dv}{dt}=-g+\frac{v^2}{1325m}

\frac{dv}{dt}=\frac{v^2 - 1325mg}{1325m}

\int \frac{dv}{v^2 - 1325mg} = \frac{1}{1325m}\int dt

Consider the left-hand side:

\int \frac{dv}{v^2 - 1325mg}=-\frac{1}{1325mg} \int \frac{1325mg}{1325mg-v^2} \: dv

\int \frac{dv}{v^2 - 1325mg}=-\frac{1}{1325mg} \int \frac{dv}{1-\left( \frac{v}{\sqrt{1325mg}} \right) ^2}

\int \frac{dv}{v^2 - 1325mg}=-\frac{1}{1325mg} \mbox{arctanh} \left( \frac{v}{\sqrt{1325mg}} \right) + \mathrm{C}

Thus, we have

-\frac{1}{1325mg} \mbox{arctanh} \left( \frac{v}{\sqrt{1325mg}} \right) = \frac{t}{1325m} + \mathrm{C}

\mbox{arctanh} \left( \frac{v}{\sqrt{1325mg}} \right) = -\frac{gt}{\sqrt{1325mg}} + \mathrm{C}

v(t) = \sqrt{1325mg} \tanh \left( -\frac{gt}{\sqrt{1325mg}} + \mathrm{C} \right)

v(0)=0 \Rightarrow \mathrm{C}=0

v(t) = \sqrt{1325mg} \tanh \left( -\frac{gt}{\sqrt{1325mg}} \right)

and so

x(t) = \sqrt{1325mg} \int \tanh \left( -\frac{gt}{\sqrt{1325mg}} \right) \: dt

The substitution

u=-\frac{gt}{\sqrt{1325mg}}

gives

x(t) = -1325m \int \tanh u \: du

x(t) = -1325m \ln (\cosh u) + \mathrm{C}

x(0)=x_{\mbox{max}} \Rightarrow \mathrm{C} = x_{\mbox{max}}

x(t) = -1325m \ln \left[ \cosh \left( \frac{gt}{\sqrt{1325mg}} \right) \right] + x_{\mbox{max}}

Now, we can find the time for the ball to fall from the maximum height

x(t)=0 \Rightarrow t \approx 3.277723004542093 \mbox{ s}

which is added to the time found in part (a)

t \approx 1.9161467809752113 \mbox{ s}

and gives the total time

t \approx 5.194 \mbox{ s}