# Reasoning/evaluation: Free fall motion with constant acceleration

• simphys
In summary, by comparing the time interval of the free fall motion(stage2) to the time needed to complete stage1, it is reasonable to infer that the time needed to complete stage1 will be less than the time needed to go from the start of free fall to the maximum altitude.
simphys
Homework Statement
A rocket is launched from the surface and reaches a maximum height of 960m. The rocket’s engine gives an 16m/s^2 upward acceleration during time T. After that time T the rocket is in free fall.

What must be the value of T in order for the rocket to reach the required altitude(of 960m)?
Relevant Equations
Getting the solution was no problem.(steps included in picture)
My question is how I could go about reasoning whether the solution of the problem makes sense without actually looking at the solution in the back.

Here f.e. T=6.75 for the ‘first’ stage, how can I reason/evaluate this without looking at the solution?
My guess is that I could compare it to the time interval of the free fall motion(stage2).
Would that be enough to let it make ‘sense’?

As we know that stage 1 has an acceleration of 16m/s^2 and free fall one of 9.81m/s^2. Then.. is it sufficient to reason that the time needed to complete stage 1 will be less than the time need to go from the start of free fall to the maximum altitude?

Welcome!
Think of fireworks.
The way up is much faster than the way down.
There is a much more abrupt change of velocity (launch to zero) during ascension (greater acceleration).

simphys
Lnewqban said:
Welcome!
Think of fireworks.
The way up is much faster than the way down.
There is a much more abrupt change of velocity (launch to zero) during ascension (greater acceleration).
Thanks! And thanks for answering as well.
Oooh okay yeah so that is the kind of reasoning to make.

So just to clarify.
In the evaluating part of a problem, I don’t really need to spend a lot of time here. It’s just about checking whether the answer makes sense as as an example, the problem above: if the ascending time would be bigger than the free fall one, it wouldn’t make sense and I’d need to revise my solution. And even after making this evaluation the answer could still be wrong but I will at least have a little certainty that it is at least a step in the right direction, correct?

Once again, thanks for the help

Lnewqban
You could also draw a graph to visualize velocities and covered height.
You are welcome.

simphys
simphys said:
My question is how I could go about reasoning whether the solution of the problem makes sense without actually looking at the solution in the back.
I'm having trouble understanding this question. If you don't look at the solution in the back, then it seems you can't go about reasoning whether the solution makes sense, because you don't know what the solution is!

If you're trying to develop the reasoning skills then you have to solve the problem without looking in the back. Then you can look in the back and compare the solution you see there with your solution. It takes practice to develop this skill but it's a skill necessary for your success.

You're taking the right steps, you just need to explain why. So, for example, you calculate that ##v_y## is 16##T##. But you don't explain that that is the velocity when the engines shut down. You can take the value of ##T## that you calculated, multiply it by 16, and see if that's a reasonable value for the speed at that time.

Hi @simphys. Welcome t PF.

Your solution looks OK - except you have written
960 – 8T
960 – 8T²
but I think that you have used the correct value ( 960 – 8T²) subsequently.

Why not simply use the value of T to calculate the final height. Not a lot of work:
##Δy_1 = 8*6.75^2##
##v_1 = 16*6.75##
##Δy_2 = v_1^2 / 2g##
Do ##Δy_1## and ##Δy_2## add up to 960m?

simphys, PeroK and Mister T
simphys said:
Here f.e. T=6.75 for the ‘first’ stage, how can I reason/evaluate this without looking at the solution?
There are lots of ways to do this problem. If you do it two different ways and get the same answer that's a good sign. Here's perhaps the wackiest way to do it.

The times ##T_1## and ##T_2## are in inverse proportion to the acceleration ##a## and ##g##, so that:
$$T_2 = \frac a g T_1$$The distance is the overall time multiplied by the average speed:
$$D = v_{avg}(T_1 + T_2) = v_{avg}(T_1 + \frac a g T_1) = v_{avg}T_1(\frac{g + a}{g})$$The average speed is the same for both legs(!), hence:$$v_{avg} = \frac 1 2 aT_1$$That gives:$$T_1^2 = \frac{2Dg}{a(a+g)}$$Wouldn't it be amazing if that gives the same answer?!

simphys and Steve4Physics
PeroK said:
There are lots of ways to do this problem
I am awed by wacky but I prefer short and simple.
For the powered leg of the journey
$$h_1=\frac{1}{2}aT_1^2$$ $$2a \left(\frac{1}{2}aT_1^2\right)=v_1^2$$For the free fall leg $$2g\left(D-\frac{1}{2}aT_1^2\right)=v_1^2=2a \left(\frac{1}{2}aT_1^2\right)$$ $$T_1^2 = \frac{2Dg}{a(a+g)}.$$

Mister T said:
I'm having trouble understanding this question. If you don't look at the solution in the back, then it seems you can't go about reasoning whether the solution makes sense, because you don't know what the solution is!

If you're trying to develop the reasoning skills then you have to solve the problem without looking in the back. Then you can look in the back and compare the solution you see there with the solution in the back. It takes practice to develop this skill but it's a skill necessary for your success.You're taking the right steps, you just need to explain why. So, for example, you calculate that ##v_y## is 16##T##. But you don't explain that that is the velocity when the engines shut down. You can take the value of ##T## that you calculated, multiply it by 16, and see if that's a reasonable value for the speed at that time.
Oh yes, most definitely! I completely agree, so let me clarify a bit. I am talking about the reasoning before looking at the solution, haha.
The way I do it is solve a problem, then look at the back and either resolve if it gets too messy or look at the mistake I made like you stated.

But suppose that you need to calculate f.e. the velocity in the real world, you can’t just look at a solution to compare it with. You’d need to use your own intuition on whether it’d be correct after making the calculation. But like you said, that takes practice..

As to why i asked the question
What Happened was.. I was just about to start kinetic energy when I realized that I don’t really think about my solutions itself(no matter wrong/correct) before looking at the Endofbook solution. Hence why I started from the beginning, and whether that is a waste of time or not at this point of time.. That I don’t really know
To be even more specific, I was doing exercises from hibbeler’s dynamics book from university which were chapters about kinematics and kinetics. I kid you not every exercise I did was wrong, but was 100% in the math, mostly how i chose the signs relative to the datum lines. Thus.. this made me skip this exam and learn it on my own with a more beginner friendly book

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Steve4Physics said:
Hi @simphys. Welcome t PF.

Your solution looks OK - except you have written
960 – 8T
960 – 8T²
but I think that you have used the correct value ( 960 – 8T²) subsequently.

Why not simply use the value of T to calculate the final height. Not a lot of work:
##Δy_1 = 8*6.75^2##
##v_1 = 16*6.75##
##Δy_2 = v_1^2 / 2g##
Do ##Δy_1## and ##Δy_2## add up to 960m?
Thanks a lot!
ooh! Yes.. thanks once again. Well.. I didn’t really think about this relation. That’s indeed the simplest one I could’ve used to check the answer. But something like thid to check the answer to make sense, do I need to do that on every excercise or not really?

PeroK said:
There are lots of ways to do this problem. If you do it two different ways and get the same answer that's a good sign. Here's perhaps the wackiest way to do it.

The times ##T_1## and ##T_2## are in inverse proportion to the acceleration ##a## and ##g##, so that:
$$T_2 = \frac a g T_1$$The distance is the overall time multiplied by the average speed:
$$D = v_{avg}(T_1 + T_2) = v_{avg}(T_1 + \frac a g T_1) = v_{avg}T_1(\frac{g + a}{g})$$The average speed is the same for both legs(!), hence:$$v_{avg} = \frac 1 2 aT_1$$That gives:$$T_1^2 = \frac{2Dg}{a(a+g)}$$Wouldn't it be amazing if that gives the same answer?!
Lol, that’d indeed amaze me That is an interesting relation that I didn’t see coming.
Thanks a lot for that view! I will check it out when I get back to it in a bit.

PeroK
simphys said:
But something like thid to check the answer to make sense, do I need to do that on every excercise or not really?
It's good practice in general. For example in maths we were taught that if you solve simultaneous equations, you should always check that the solution satisfies the equations.

But it's down to context and preferences. E.g. for an exam, it might be best use of time to leave checking till after you've answered all questions (if there is any time left).

For homewotk or designing a safety-critical nuclear reactor system, checking seems the best option.

simphys
kuruman said:
I am awed by wacky but I prefer short and simple.
For the powered leg of the journey
$$h_1=\frac{1}{2}aT_1^2$$ $$2a \left(\frac{1}{2}aT_1^2\right)=v_1^2$$For the free fall leg $$2g\left(D-\frac{1}{2}aT_1^2\right)=v_1^2=2a \left(\frac{1}{2}aT_1^2\right)$$ $$T_1^2 = \frac{2Dg}{a(a+g)}.$$
Haha I am also a fan of that.
But hey, that looked so simple solving algebraically! And more error prone(?) i always plug and chug at the beginning/middle making lots mistakes along way to the end…
Thanks by the way!

Steve4Physics said:
It's good practice in general. For example in maths we were taught that if you solve simultaneous equations, you should always check that the solution satisfies the equations.

But it's down to context and preferences. E.g. for an exam, it might be best use of time to leave checking till after you've answered all questions (if there is any time left).

For homewotk or designing a safety-critical nuclear reactor system, checking seems the best option.
Argh self-taught the math so I don’t really doit
And haha most definitely important in the design.
But thanks a lot, great point! I’ll just move a bit quicker through the exercises then.

The simplest solution is to draw a graph of speed against time, where the total area is the distance ##D##. There's a lot that you can see straight from the graph:

From the graph it's obvious the speed at the end of the acceleration phase is the starting speed at the beginning of the deceleration phase. I.e. $$v_{max} = aT_1 = gT_2$$It also shows the symmetry that the deceleration phase with ##g## has the same shape and area (distance) as would an acceleration for the same time ##T_2## with positive ##g##.

You can also see that the average height of each triangle is the same: it's half the maximum height. I.e. both legs have the same average speed, which is half the maximum speed:$$v_{avg} = \frac 1 2 v_{max} = \frac 1 2 aT_1 = \frac 1 2 gT_2$$And, of course, the total distance ##D## is given by the sum of the areas of the two triangles, which is ##\frac 1 2 \times \text{base} \times \text{height}##:$$D = \frac 1 2 T_1(aT_1) + \frac 1 2T_2(gT_2) = \frac 1 2 aT_1^2 + \frac 1 2 gT_2^2$$To solve the problem, we just need to substitute ##T_2 = \frac a g T_1## into that equation.

Lnewqban and simphys
PeroK said:
The simplest solution is to draw a graph of speed against time, where the total area is the distance ##D##. There's a lot that you can see straight from the graph:

From the graph it's obvious the speed at the end of the acceleration phase is the starting speed at the beginning of the deceleration phase. I.e. $$v_{max} = aT_1 = gT_2$$It also shows the symmetry that the deceleration phase with ##g## has the same shape and area (distance) as would an acceleration for the same time ##T_2## with positive ##g##.

You can also see that the average height of each triangle is the same: it's half the maximum height. I.e. both legs have the same average speed, which is half the maximum speed:$$v_{avg} = \frac 1 2 v_{max} = \frac 1 2 aT_1 = \frac 1 2 gT_2$$And, of course, the total distance ##D## is given by the sum of the areas of the two triangles, which is ##\frac 1 2 \times \text{base} \times \text{height}##:$$D = \frac 1 2 T_1(aT_1) + \frac 1 2T_2(gT_2) = \frac 1 2 aT_1^2 + \frac 1 2 gT_2^2$$To solve the problem, we just need to substitute ##T_2 = \frac a g T_1## into that equation.
Wow thanks, I really appreciate it! I didn't realize such a solution was possible. very interesting.
question. The way you laid out this solution, to which kind of problems is this applicable then? Or just in general drawing the graph and seeing what kind of connection you'd be able to make?

simphys said:
Wow thanks, I really appreciate it! I didn't realize such a solution was possible. very interesting.
question. The way you laid out this solution, to which kind of problems is this applicable then? Or just in general drawing the graph and seeing what kind of connection you'd be able to make?
For constant acceleration problems, drawing a graph of velocity against time is an underused technique. Things are much clearer from the graph and this often provides a quick way to see what equations to use, even if you do then turn to algebra.

And, especially if you have acceleration followed by constant velocity following by deceleration (or something like that), it let's you divide up the equations of motion with minimal algebraic trouble.

For variable acceleration, it's just as useful, although there we are talking about integrating the velocity function to get the displacement. I.e. it's not just triangles and rectangles.

It can also be useful to resolve conceptual ideas. E.g. some people think that if a ball is thrown up, then at the top of the trajectory (when the velocity is zero) then the acceleration must be zero. Drawing a velocity against time graph shows that the slope of the line is constant (from ##+u## to ##0## to ##-u##) and that, in fact, the point where ##v = 0## has no special significance regarding gravitational acceleration.

Lnewqban and simphys
PS also think about using a displacement against time graph for interception problems! Students tend to go round in circles with various SUVAT formulas, where a displacement against time for two vehicles shows much more clearly where and when one vehicle catches and overtakes another.

Lnewqban
PeroK said:
For constant acceleration problems, drawing a graph of velocity against time is an underused technique. Things are much clearer from the graph and this often provides a quick way to see what equations to use, even if you do then turn to algebra.

And, especially if you have acceleration followed by constant velocity following by deceleration (or something like that), it let's you divide up the equations of motion with minimal algebraic trouble.

For variable acceleration, it's just as useful, although there we are talking about integrating the velocity function to get the displacement. I.e. it's not just triangles and rectangles.

It can also be useful to resolve conceptual ideas. E.g. some people think that if a ball is thrown up, then at the top of the trajectory (when the velocity is zero) then the acceleration must be zero. Drawing a velocity against time graph shows that the slope of the line is constant (from ##+u## to ##0## to ##-u##) and that, in fact, the point where ##v = 0## has no special significance regarding gravitational acceleration.
Wooww, amazing! Underused is an understatement, if the graph wasn't given I wouldn't even bother making one :'D (ofcourse besides a FBD but then we're talking forces).
I always just think about it in my head, like oh yeah ymax, certainly v=0 then, but f.e. in free fall still the same acceleration. on paper on the other hand it'd be a matter of just reading it out.
Once again thanks a lot for the useful information!

And I also notice how you keep it in algebra all the way through to the end, will start doing that too hehe

berkeman and PeroK
PeroK said:
It can also be useful to resolve conceptual ideas. E.g. some people think that if a ball is thrown up, then at the top of the trajectory (when the velocity is zero) then the acceleration must be zero. Drawing a velocity against time graph shows that the slope of the line is constant (from ##+u## to ##0## to ##-u##) and that, in fact, the point where ##v = 0## has no special significance regarding gravitational acceleration.
And if the ball is thrown up in the more realistic case of including air resistance, it is useful in figuring out without equations which leg takes longer time, going up or coming back down. The slopes are different but the areas under the curve are the same.

Lnewqban, simphys and PeroK
kuruman said:
And if the ball is thrown up in the more realistic case of including air resistance, it is useful in figuring out without equations which leg takes longer time, going up or coming back down. The slopes are different but the areas under the curve are the same.
if I am not mistaken air restistance was proportional to the speed so according to my thinking the air resistance will be the biggest when the speed is the biggest in magnitude (and for long distances (or heavy objects idk anymore) it was proportional to the square of the speed I think)

simphys said:
if I am not mistaken air restistance was proportional to the speed so according to my thinking the air resistance will be the biggest when the speed is the biggest in magnitude (and for long distances (or heavy objects idk anymore) it was proportional to the square of the speed I think)
You don't need to go into such detail for a qualitative explanation. On the way up the acceleration has magnitude greater than ##g##; on the way down, the acceleration has magnitude less than ##g##. That makes the "down" time greater than the "up" time no matter what the model for air resistance because the area under the curve has to be the same.

Say you take a video of the entire motion and you split it in two parts, "up" and "down". Then you run the "up" video backwards. You will see the object falling from rest with "down" acceleration ##g+a(v)## from height ##h##. When you run the "down" video as is, you will also see the object falling from rest but with "down" acceleration ##g-a(v)## from the same height ##h##. Which video will run longer?

simphys
simphys said:
I don’t really think about my solutions itself(no matter wrong/correct) before looking at the Endofbook solution. Hence why I started from the beginning, and whether that is a waste of time or not at this point of time.. That I don’t really know
Thinking about the validity of a solution is not a waste of time. On the contrary, trying to learn physics without thinking about the validity of your solutions is a waste of time. You won't do well in the class and you won't learn any physics.

jbriggs444
Mister T said:
Thinking about the validity of a solution is not a waste of time. On the contrary, trying to learn physics without thinking about the validity of your solutions is a waste of time. You won't do well in the class and you won't learn any physics.
Yes! Well.. that is why I started all over again to approach it in such a way, really makes a difference. (btw the don't think needed to be didn't think (past) :) )

I was wondering if it was possible to answer me one more question related to math --> I really need some bigtime advice on the math part. (I know it is not the right thread but my sincere apologies)
So what you apply in physics of math, do you (implicitly perhaps) use geometry in solving it or some kinda intuition developed there? Cause I really feel that I miss some kind of outside part if you know what I mean.. but hey that might be me aswell...
I started studying math 6-7 months ago, but with precalc and a bit of calc and then came calc 1 at university and essentially skipped geometry. And for high school.. in general the Math part was bad and didn't get physics.(am 19 atm)

The biggest hurdle is that I keep questioning the geometry part which implies that I START to question all my accumulated math skills those last couple months. Like oh man shouldn't I also go over precalc again this and that. Again headgames.

Now to ask more specific toward the action part as I can keep whining all day long but without action no change's bound to happen...
What would be the better approach.. stopping uni right now, learn geometry,... and start all over again next year , or just simply do great time management and fill it in with geometry learning in the meantime if that is even needed and perhaps going over a good algebra book after that to really make sure the foundations are solid? And also catch up with the analytical thinking that Physics provides. (ohyeah I study engineering so.. intro physics this semester was basically more practical with no conceptual explanation at all)
or.. would you recommended to approach it a different way?

Thanks in advance, again my apologies for dropping all my doubts and concerns into this thread, but I didn't have anyone to really talk with or that'd be able to give me advice on it until I found this amazing forum!

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simphys said:
Now to ask more specific toward the action part as I can keep whining all day long but without action no change's bound to happen...
You are welcome to ask that question but not here. This forum is specifically for Introductory physics homework. I recommend that you start a new thread in forum STEM Academic Advising.

simphys
m
kuruman said:
You are welcome to ask that question but not here. This forum is specifically for Introductory physics homework. I recommend that you start a new thread in forum STEM Academic Advising.
Thanks my apologies once again

simphys said:
So what you apply in physics of math, do you (implicitly perhaps) use geometry in solving it or some kinda intuition developed there?
You explicitly use geometry when you study physics or engineering.

simphys said:
What would be the better approach.. stopping uni right now, learn geometry,... and start all over again next year , or just simply do great time management and fill it in with geometry learning in the meantime

I recommend the latter. I suspect you are being a bit of a perfectionist.

kuruman said:
You are welcome to ask that question but not here. This forum is specifically for Introductory physics homework. I recommend that you start a new thread in forum STEM Academic Advising.

Mister T said:
You explicitly use geometry when you study physics or engineering.
I recommend the latter. I suspect you are being a bit of a perfectionist.
ow interesting, that is very great to know.

Will follow through like that then, thank you! And yep you are right I was. but learned a lot about how wrong I was approaching it and that It's important that I need to devellop a valid study strategy as well.

I am going to open one right now.
Thanks!

## 1. What is free fall motion with constant acceleration?

Free fall motion with constant acceleration is a type of motion where an object is falling under the influence of gravity with a constant acceleration. This means that the object's velocity increases by the same amount every second.

## 2. What is the acceleration due to gravity?

The acceleration due to gravity is the acceleration that an object experiences when it is falling towards the Earth due to the force of gravity. On Earth, this value is approximately 9.8 meters per second squared (m/s²).

## 3. How is free fall motion with constant acceleration different from other types of motion?

Free fall motion with constant acceleration is different from other types of motion because the acceleration is constant, meaning it does not change over time. In other types of motion, the acceleration may vary depending on different factors.

## 4. What is the formula for calculating the distance traveled during free fall motion with constant acceleration?

The formula for calculating the distance traveled during free fall motion with constant acceleration is d = (1/2)at², where d is the distance, a is the acceleration, and t is the time.

## 5. How is free fall motion with constant acceleration used in real-life applications?

Free fall motion with constant acceleration is used in many real-life applications, such as skydiving, bungee jumping, and amusement park rides. It is also used in physics experiments and calculations, as well as in the design of objects that need to be dropped or fall from a certain height.

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