Maximum Duration of a Projectile Motion Flight with Air Resistance

[timloser1987]

Homework Statement

A rock is launched vertically.
During the last second of the flight, the rock covers one-half of the entire distance covered during the entire flight. What is the maximum possible duration of the flight?

Hint: Answer $\neq 2 s$

Homework Equations

Consider 1-D motion:

When the rock is moving upwards,

$$-(mg+b\dot{y})=m\ddot{y}$$.

When the rock is falling,

$$-(c\dot{y}-mg)=m\ddot{y}$$

The Attempt at a Solution

I've solved for the velocity and displacement functions but somehow I kept getting 2 seconds as my answer. The question isn't very explicit in stating what you should consider, but I assumed it to be air resistance that is directly proportional to the velocity $R=-kv$. Intuitively it would appear that for any air resistance in order to maximize the duration of the flight the resistance constant $k$ will have to be zero. But somehow it just doesn't work that way.

Is there something wrong with my assumption that air resistance is proportional to $v$ and not $v^{2}$, since I've also learned that it can be proportional to the square of the velocity, too.

Some guy suggested that this can be modeled such that the gravitational force is non-constant, i.e. use $F_{g}=\frac{Gm_{1}m_{2}}{r^2}$, but I have no idea how to solve that way.

Can someone please help? Thanks.

 

[anathema]

Thank you for your post. I would like to offer some insight on your question.

First, it is important to note that air resistance can indeed be modeled using both linear and quadratic functions of velocity. However, for this particular problem, we will assume that air resistance is negligible.

Now, let's consider the motion of the rock. From the given information, we know that during the last second of the flight, the rock covers half of the total distance. This means that during the first part of the flight, it covers the other half of the distance. Therefore, we can say that the rock reaches its maximum height at the midpoint of the flight.

Using this information, we can set up the following equation:

$$y_{max}=\frac{1}{2}gt^2$$

Where $y_{max}$ is the maximum height reached by the rock and $t$ is the total duration of the flight.

Now, we also know that the velocity at the top of the flight is zero, since the rock momentarily stops before falling back down. This means that at the midpoint of the flight, the velocity is also zero. Using this information, we can set up the following equation:

$$v_{mid}=gt$$

Where $v_{mid}$ is the velocity at the midpoint of the flight.

Now, we can solve for $t$ by setting these two equations equal to each other and solving for $t$:

$$\frac{1}{2}gt^2=gt$$

$$t=2$$

As you can see, we get the same answer of 2 seconds. This is because we are assuming that air resistance is negligible, so the equations you have set up are correct. Therefore, the maximum possible duration of the flight is indeed 2 seconds.

I hope this helps clarify your understanding of the problem. If you have any further questions, please don't hesitate to ask.

 

[baba89]

First of all, I would like to commend you for considering air resistance in your solution. It is important to take into account all relevant factors in order to accurately model the projectile motion.

Now, let's take a closer look at your equations. The first equation you have written for when the rock is moving upwards is correct, but the second equation for when the rock is falling is incorrect. The correct equation should be:

$$-(mg+c\dot{y})=m\ddot{y}$$

This is because when the rock is falling, the air resistance is acting in the opposite direction of the velocity, hence the negative sign in front of the air resistance term.

Now, to address your question about the air resistance being proportional to $v$ or $v^2$, both assumptions are valid depending on the specific situation. In this case, since you are considering a simple projectile motion with no other external forces acting on the rock, it would be more accurate to model the air resistance as being directly proportional to the velocity. However, if there were other external forces present, such as wind, then the air resistance term would be proportional to the square of the velocity.

As for the maximum duration of the flight, it cannot be simply determined by setting the air resistance constant $k$ to zero. The maximum duration of the flight will depend on the initial velocity and the air resistance constant $k$. It is important to note that as the initial velocity increases, the air resistance will also increase, thus decreasing the maximum duration of the flight.

Lastly, the suggestion to use the gravitational force equation $F_{g}=\frac{Gm_{1}m_{2}}{r^2}$ is not necessary in this case. This equation is used for calculating the gravitational force between two objects, and it is not relevant in a simple projectile motion scenario.

To solve for the maximum duration of the flight, you can use the equations you have derived for the velocity and displacement, and set the displacement equal to one-half of the total distance covered. This will give you the time at which the rock reaches its maximum height, and from there you can calculate the maximum duration of the flight.

I hope this helps clarify some points and guides you towards the correct solution. Good luck!