Height and freefalling question

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In summary, according to the student, the book gives three different answers for the height of a body in freefall. The teacher told the student that it might be a mistake in the task by the authors, but if they find a solution to the book's answers then he will give them a bonus on the test. The student tried looking on the internet for a solution but could not find one. The student then came to Physics Forums and asked for help.
  • #1
Gustavs1337
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There was a problem with other physics homework section so I just decided to write here.

Homework Statement


H= 270m t1= 2.45s t3=7.35 V0=0 (freefall)
t=7.35s t2= 4.90s g=10
We had to calculate the height what the body will be after 2.45, 4.90 and 7.35 s.

Homework Equations


h=gt2/2

The Attempt at a Solution


h1=30 h2= 120 h3= 270

So this is what it was supposed to look like. But, the answers in the book say that h1=30 h2=90 and h3= 150. Our teacher told us that it might be a mistake in the task by the authors. But he said that if we find a solution to the books answers then he will give the person +2 on the test ( this was a test task btw). So i could not think of anything since it does not make sense. I tried looking on the internet and i found this forum. If you have any ideas, please share them.
 
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  • #2
Gustavs1337 said:
There was a problem with other physics homework section so I just decided to write here.

Homework Statement


H= 270m t1= 2.45s t3=7.35 V0=0 (freefall)
t=7.35s t2= 4.90s g=10
We had to calculate the height what the body will be after 2.45, 4.90 and 7.35 s.

Homework Equations


h=gt2/2

The Attempt at a Solution


h1=30 h2= 120 h3= 270

So this is what it was supposed to look like. But, the answers in the book say that h1=30 h2=90 and h3= 150. Our teacher told us that it might be a mistake in the task by the authors. But he said that if we find a solution to the books answers then he will give the person +2 on the test ( this was a test task btw). So i could not think of anything since it does not make sense. I tried looking on the internet and i found this forum. If you have any ideas, please share them.
Hi Gustavas, Welcome to Physics Forums.

None of the answers make much sense. If the body is truly in freefall from rest (zero initial velocity) then its height should be decreasing with time.
 
  • #3
If it's free falling then the height will decrease with time. You seem to have an object accelerating upwards from 0.
 
  • #4
Yea its supposed to be falling, and it is, maybe you got the wrong idea, because of what i wrote. We have to calculate where will it be ( how much would it fall) after the 3 same periods of time ( the period is 2.45 sec). So it falls 30 m the first 2.45 sec, 120m 4.90 s, and its 270 when 7.35 s. But the answers in the book say different ( 30,90,150). If the answers in the book are impossible if it was free falling then it is impossible. The teacher himself couldn't figure it out. We get bonus if we figure it out.
 
  • #5
If your teacher can't figure that one out, there must be something you're not telling us! Figure what out? That the book has given the answer to a different question, perhaps.
 
  • #6
The given answers would make sense if the question asked for the distance covered by the falling body between the given instants of time. So for example from time zero to time 2.45 seconds the abject falls a distance of 30 meters.
 
  • #7
Figure out why it is 30, 90 , 150 in the book but with the formulas it is 30, 120, 270. I already told you that it might be impossible, just a mistake. I would show you but you are probably not latvian. And the answers are to that task. And also if the body was freefalling it would have acceleration, so it cannot be 30, 90, 150. I do not think that this kind of mistake would be in the book. But there might be some physics things to it. I don't really know that's why I am asking.
 
  • #8
I disagree with your conclusion. My calculations show 30, 90, and 150 meters for the given spans. You'll have to prove me wrong :smile:

Write an expression for the distance traveled with respect to time and find the difference between two given times.
 
  • #9
Can you show me how you calculated that please? This is my calculation: gt^2/2= 10x2.45^2/2 = 30, h2= 10x4.90^2=120 , h3= 10x7.35^2=270. How can it be 30 90 and 150 for you. There would be an acceleration and the differencebetween 30, 90 and 150 is 60 meters. Please explain. I could not understand what do i have to reply because english is not my native language. ''Write an expression for the distance traveled with respect to time and find the difference between two given times.''
 
  • #10
Write an equation for the distance with respect to time. Use that equation to find the total distance traveled (from rest) at each of the given times. What values do you find?
 
  • #11
It is going to be hard to understand for me, since I am on the verge of failing physics. The equation you mean like Distance/Time for example m/t? I also might understand something from your calculation of this. If you do not want to share it and you have the time to explain then I will stick here too.
 
  • #12
If a body starts from rest with an acceleration a, what is an equation that express the distance with respect to time?
 
  • #13
is it a=deltaV/t or do you mean a=v-v0/t. I am sorry if I am being a bother.
 
  • #14
Gustavs1337 said:
is it a=deltaV/t or do you mean a=v-v0/t. I am sorry if I am being a bother.
You need to review your basic kinematic equations. What is the equation of motion for a body starting from rest with a given acceleration? d = ?
 
  • #15
oh you mean d=v0t+1/2at^2? or
 
  • #16
Gustavs1337 said:
oh you mean d=v0t+1/2at^2? or
Yes.

So if it starts from rest and has an acceleration of 10 m/s2, what are the distances at 0s, 2.45s, 4.90s, and 7.35s?
 
  • #17
yes i know now g=a don't answer that. 0 would be 270, 2.45 is ~30m, the next i don't understand. if a is 10m/s2 then d= 0x4.90+10x4.90^2/2= 120 and d=0x7.35+10x7.35^2/2 = 270
https://gyazo.com/320ae67296b95a7ec9f69170093ccf51 :H:frown:. I can't make it 30,90,150
 
  • #18
Gustavs1337 said:
yes i know now g=a don't answer that. 0 would be 270, 2.45 is ~30m, the next i don't understand. if a is 10m/s2 then d= 0x4.90+10x4.90^2/2= 120 and d=0x7.35+10x7.35^2/2 = 270
https://gyazo.com/320ae67296b95a7ec9f69170093ccf51 :H:frown:. I can't make it 30,90,150
You're saying that at time zero the object has traveled 270 m? How is that possible if it hasn't moved yet? At time zero it has traveled zero meters. We aren't discussing height anymore, so the initial height of 270 m is irrelevant for now.

So your times and distances are:

t0 = 0s ; d0 = 0 m
t1 = 2.45 s ; d1 = 30 m
t2 = 4.90 s ; d2 = 120 m
t3 = 7.35 s ; d3 = 270 m

Those are the total distances (displacements) from the starting point for each of those particular times. Now, what are the distances traveled between those instants of time? For example, what is the distance covered from time t2 to time t3?
 
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  • #19
Omg i now understand, its 30, 90 ,150. Well i don't understand to the fullest i might use this thread in the future. Thanks you are a great help. I lack logic.
 

1. What is the relationship between height and freefalling?

The relationship between height and freefalling is that the higher the initial height, the longer the object will take to reach the ground. This is because the object has more distance to cover before reaching the ground, and gravity will cause it to accelerate towards the ground at a constant rate.

2. How does air resistance affect freefalling from a high height?

Air resistance, also known as drag, can affect freefalling from a high height by slowing down the object's descent. This is because as the object falls, it will encounter air molecules, which will push against it in the opposite direction of its motion, causing it to slow down. The higher the height, the more air resistance it will experience, and the longer it will take to reach the ground.

3. What is terminal velocity and how does it relate to freefalling?

Terminal velocity is the maximum speed that an object can reach while falling due to the balance between gravity and air resistance. As an object falls from a high height, it will accelerate due to gravity, but as it picks up speed, air resistance will also increase, eventually balancing out the acceleration. This results in the object reaching a constant speed known as terminal velocity, which it will maintain until it reaches the ground.

4. What are the factors that affect the height and time of freefalling?

The two main factors that affect the height and time of freefalling are the initial height and the acceleration due to gravity. The higher the initial height, the longer the object will take to fall, and the faster it will be traveling when it reaches the ground. The acceleration due to gravity, which is a constant 9.8 m/s², also plays a significant role in determining the object's speed and time of freefalling.

5. Can an object ever reach a height of zero when freefalling?

No, an object cannot reach a height of zero when freefalling unless there is an external force acting on it, such as a parachute opening. This is because, in freefall, an object will continue to accelerate towards the ground at a constant rate until it reaches the ground, and at that point, its height will be zero. However, if there is no external force, the object will continue to bounce back up after reaching the ground due to the force of its impact, never reaching a height of zero.

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