First-order differential equation

AI Thread Summary
A ball with a mass of 0.15 kg is thrown upward with an initial velocity of 20 m/s from a 30 m high building, facing air resistance proportional to its velocity. The motion is analyzed in two phases: ascent and descent, using a first-order differential equation to account for forces acting on the ball. The correct time for the ball to hit the ground is calculated to be approximately 5.129 seconds, despite initial confusion regarding the differential equation's terms. The discussions clarify that the negative sign in front of the gravitational force is appropriate as it indicates the direction of acceleration opposing the motion. Overall, the problem emphasizes the importance of correctly interpreting the forces involved in projectile motion.
DivGradCurl
Messages
364
Reaction score
0
"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:
Physics news on Phys.org
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.
 
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
 
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
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top