# First-order differential equation for ball of mass

• DivGradCurl
In summary, the conversation discusses the first-order differential equation set up based on the given information of a ball being thrown upward with initial velocity 20 m/s from the roof of a building 30 m high. The equation includes the force due to air resistance, which is represented by \frac{v^2}{1325}, where velocity v is measured in m/s. The conversation also explores the use of integrating factors to solve the equation, and suggests using the substitution v=\sqrt{1325mg}\tan \theta. It is later discovered that the quadratic air resistance law should be -k|v|v, where |v| is the speed of the object. This is used to find the correct solution
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.

HINT:Variables are separable.

Daniel.

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

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

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

Last edited:
Yes,it looks okay.

Daniel.

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

Last edited:
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

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

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

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

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

Last edited:
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.

Last edited:
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

$$t \approx 5.194 \mbox{ s}$$

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## 1. What is a first-order differential equation for a ball of mass?

A first-order differential equation for a ball of mass is an equation that describes the motion of a ball based on its mass, velocity, and the external forces acting upon it. It is a mathematical representation of Newton's second law of motion.

## 2. How is the first-order differential equation for a ball of mass derived?

The first-order differential equation for a ball of mass is derived using the principles of calculus and Newton's second law of motion. It involves taking the derivative of the ball's position function with respect to time and setting it equal to the sum of all external forces acting on the ball, divided by its mass.

## 3. Can the first-order differential equation for a ball of mass be solved analytically?

Yes, the first-order differential equation for a ball of mass can be solved analytically using techniques such as separation of variables or integrating factors. However, in some cases, it may not be possible to find an exact solution and numerical methods may be used instead.

## 4. What are the applications of the first-order differential equation for a ball of mass?

The first-order differential equation for a ball of mass has various applications in physics and engineering, such as predicting the trajectory of a projectile, analyzing the motion of objects in free fall, and designing control systems for objects in motion.

## 5. Are there any limitations to the first-order differential equation for a ball of mass?

Yes, the first-order differential equation for a ball of mass is a simplified model that does not take into account factors such as air resistance, friction, and the shape of the ball. In real-world scenarios, these factors can significantly impact the motion of the ball and may need to be considered in the equation.

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