# First-order differential equation

1. Jun 9, 2005

"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 $$\left| v \right|/30$$ where the velocity $$v$$ 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 $$x$$ positively upward from the ground, then

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

$$\frac{dv}{dt} + \frac{v}{30m}=-g$$

The Method of Integrating Factors gives

$$\mu = \exp \left( \frac{1}{30m} \int dt \right) =\exp \left( \frac{t}{30m} \right)$$

$$v(t)=\exp \left( -\frac{t}{30m} \right) \int -g\exp \left( \frac{t}{30m} \right) \: dt$$

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

$$v(t)=-30gm + \mathrm{C} \exp \left( -\frac{t}{30m} \right)$$

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

$$v(0) = v_0 \Rightarrow \mathrm{C} = v_0 + 30 gm$$

Thus, we obtain

$$v(t) = -30gm + \left( v_0 + 30 gm \right) \exp \left( -\frac{t}{30m} \right)$$

At the maximum height, we have

$$\frac{dx}{dt}=v=0$$

which gives

$$t_{\mbox{up}} = 30m \ln \left( \frac{v_0}{30gm} + 1 \right)$$

The way down is described as follows:

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

$$\frac{dv}{dt} - \frac{v}{30m}=-g$$

The Method of Integrating Factors gives

$$\mu = \exp \left( -\frac{1}{30m} \int dt \right) =\exp \left( -\frac{t}{30m} \right)$$

$$v(t)=\exp \left( \frac{t}{30m} \right) \int -g \exp \left( -\frac{t}{30m} \right) \: dt$$

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

$$v(t)= 30mg + \mathrm{C}\exp \left( \frac{t}{30m} \right)$$

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

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

Thus, we obtain

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

and so

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

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

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

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

Thus, we obtain

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

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

$$v(t) = -30gm + \left( v_0 + 30 gm \right) \exp \left( -\frac{t}{30m} \right)$$

gives

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

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

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

$$x(0) = x_0 \Rightarrow \mathrm{C} = x_0 + 30m \left( v_0 + 30gm \right)$$

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

Again, setting $$x(t)=0$$ and solving for $$t$$ gives $$t \approx 5.129 \mbox{ s}$$. 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. Jun 9, 2005

### Galileo

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

$$ma=-mg-v/30$$
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.

3. Jun 9, 2005

Oh, sure! The reason why I posted this question is because the other day I was solving a similar DE where there was a $$v^2$$ term. It's clear to me now.

Thanks

4. Sep 6, 2011

### TheIconoclast

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?

Thanks