First-order differential equation

[DivGradCurl]

"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.

 

[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

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

 

[DivGradCurl]

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

 

[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

 

[rwisz]

Thank you.

Your calculations are correct, and the answer of 5.129 seconds is indeed the correct time for the ball to hit the ground. The reason why your second approach using the position equation x(t) gives the correct answer is because it takes into account the initial position of the ball, which is 30 meters above the ground. In your first approach, you only considered the velocity equation and did not take into account the initial position, which is why you got the wrong answer.

It may seem strange that the differential equation for position is not necessary, but this is because the initial conditions are already taken into account when solving the differential equation for velocity. The differential equation for position only becomes necessary when the initial position is not given. In this case, we already know the initial position, so it is not necessary to use the position equation.

In summary, both approaches are correct, but the second approach is more straightforward and takes into account the initial position of the ball. Keep up the good work with your calculations and problem solving!