# Newtonian Physics

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#1
We have an object that is raised 45m above the ground. It takes 1.8s for it to fall from rest to the ground. The mass of the object is 90kg. What is the resultant force by which the object hits the ground? Assume air resistance and friction is negligible.
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2 years ago
#2
(Original post by jmushtaq)
We have an object that is raised 45m above the ground. It takes 1.8s for it to fall from rest to the ground. The mass of the object is 90kg. What is the resultant force by which the object hits the ground? Assume air resistance and friction is negligible.
What have you tried so far?
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2 years ago
#3
(Original post by jmushtaq)
We have an object that is raised 45m above the ground. It takes 1.8s for it to fall from rest to the ground. The mass of the object is 90kg. What is the resultant force by which the object hits the ground? Assume air resistance and friction is negligible.
I think u use suvat equations to find acceleration. U have s=45, u=0, t= 1.8 a=?
Then resultant force = mass × acceleration.
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2 years ago
#4
(Original post by jmushtaq)
We have an object that is raised 45m above the ground. It takes 1.8s for it to fall from rest to the ground. The mass of the object is 90kg. What is the resultant force by which the object hits the ground? Assume air resistance and friction is negligible.

There is missing info in the question. When the object hits the ground with certain speed, would the object rebounded and what is the contact time between the ground and object?
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#5
(Original post by Relentas)
What have you tried so far?
This is what I've done:

s=ut+(at^2/2)
a= 2s/t^2
a= 2 x 45 x 1.8^2
a= 292 m/s^2
F=ma
F= 90 x 292
F= 26.3 kN

GPE=mgh
GPE= 90 x 9.81 x 45
GPE= 3.97 x 10^4 J
W= Fx
F= W/x
F= 3.97 x 10^4 / 45
F= 883 N

I don't know which one of us is correct or if both our us are incorrect.
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2 years ago
#6
(Original post by jmushtaq)
This is what I've done:

s=ut+(at^2/2)
a= 2s/t^2
a= 2 x 45 x 1.8^2
a= 292 m/s^2
F=ma
F= 90 x 292
F= 26.3 kN

GPE=mgh
GPE= 90 x 9.81 x 45
GPE= 3.97 x 10^4 J
W= Fx
F= W/x
F= 3.97 x 10^4 / 45
F= 883 N

I don't know which one of us is correct or if both our us are incorrect.
I would say both are incorrect, as there are missing info.
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#7
(Original post by Eimmanuel)
I would say both are incorrect, as there are missing info.
Can you please explain why the two methods are incorrect?
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2 years ago
#8
(Original post by jmushtaq)
This is what I've done:

s=ut+(at^2/2)
a= 2s/t^2
a= 2 x 45 x 1.8^2
a= 292 m/s^2
F=ma
F= 90 x 292
F= 26.3 kN

GPE=mgh
GPE= 90 x 9.81 x 45
GPE= 3.97 x 10^4 J
W= Fx
F= W/x
F= 3.97 x 10^4 / 45
F= 883 N

I don't know which one of us is correct or if both our us are incorrect.
Ok, Imma put this out there.
You know the velocity of its fall at the beginning is 25ms-1. (v=d/t, v=45/1.8)
You want that to be converted to acceleration. so div by time again. (a=v/t, a=25/1.8 = 13.8888..)
A= 13.888888ms-2......
which is also equal to the gravity.
(The acceleration isn't 9.8ms-2. - doesn't specify that you're on earth or what the gravity is, it asks you to calculate it.)

From there I assume you do what you did, sub that in.
F=ma
F=90*13.8888888 = 1250N of force.
My working to me looks right, do you agree or am I missing something, or are we both missing something? :P

Edit: I'm sure that's right. It has to be imo.
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#9
(Original post by Relentas)
Ok, Imma put this out there.
You know the velocity of its fall at the beginning is 25ms-1. (v=d/t, v=45/1.8)
You want that to be converted to acceleration. so div by time again. (a=v/t, a=25/1.8 = 13.8888..)
A= 13.888888ms-2......
which is also equal to the gravity.
(The acceleration isn't 9.8ms-2. - doesn't specify that you're on earth or what the gravity is, it asks you to calculate it.)

From there I assume you do what you did, sub that in.
F=ma
F=90*13.8888888 = 1250N of force.
My working to me looks right, do you agree or am I missing something, or are we both missing something? :P

Edit: I'm sure that's right. It has to be imo.
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2 years ago
#10
(Original post by jmushtaq)
If it was on earth then no, it wouldn't be correct. Unless the person chucked the ball/threw it downwards.
The gravity on earth varies up to around 0.5%, so it's stronger/lighter in some places, 9.8+/- 0.5% <> 14.
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2 years ago
#11
(Original post by jmushtaq)
Can you show the question in full? It doesn't say it's on earth, at least from what you've wrote on the thread.
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#12
(Original post by Relentas)
If it was on earth then no, it wouldn't be correct. Unless the person chucked the ball/threw it downwards.
The gravity on earth varies up to around 0.5%, so it's stronger/lighter in some places, 9.8+/- 0.5% <> 14.
Would it just be F=mg then assuming g is 9.8?
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2 years ago
#13
(Original post by jmushtaq)
Would it just be F=mg then assuming g is 9.8?
It must say on the question that it's on earth/what the gravity is, if not then you would be required to calculate it.
In all sciences the never give you info that you don't need - (for the most part at least.)
1
#14
(Original post by Relentas)
It must say on the question that it's on earth/what the gravity is, if not then you would be required to calculate it.
In all sciences the never give you info that you don't need - (for the most part at least.)
Okay, thanks for the help 0
2 years ago
#15
(Original post by jmushtaq)
This is what I've done:

s=ut+(at^2/2)
a= 2s/t^2
a= 2 x 45 x 1.8^2
a= 292 m/s^2
F=ma
F= 90 x 292
F= 26.3 kN

GPE=mgh
GPE= 90 x 9.81 x 45
GPE= 3.97 x 10^4 J
W= Fx
F= W/x
F= 3.97 x 10^4 / 45
F= 883 N

I don't know which one of us is correct or if both our us are incorrect.
(Original post by jmushtaq)
Can you please explain why the two methods are incorrect?

(Original post by jmushtaq)
This is what I've done:

s=ut+(at^2/2)

a= 2s/t^2

a= 2 x 45 x 1.8^2

a= 292 m/s^2

F=ma

F= 90 x 292

F= 26.3 kN

….

45 m is the distance the object falls not the distance that the object has “fallen” when it is in contact with the ground. The same reason is for your friend’s answer. But there is another reason.

(Original post by jmushtaq)
…..

GPE=mgh

GPE= 90 x 9.81 x 45

GPE= 3.97 x 10^4 J

W= Fx

F= W/x

F= 3.97 x 10^4 / 45

F= 883 N

I don't know which one of us is correct or if both our us are incorrect.

Another reason is that 883 N is just the weight of the object (W = mg = 90 × 9.81 = 883 N)
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2 years ago
#16
(Original post by jmushtaq)
(Original post by jmushtaq)
We have an object that is raised 45m above the ground. It takes 1.8s for it to fall from rest to the ground. The mass of the object is 90kg. What is the resultant force by which the object hits the ground? Assume air resistance and friction is negligible.

As I mention there are missing info or the phrasing of the question seems a bit odd and it seems that you refuse to accept it

I would elaborate further.
I assume that the object in on Earth surface.
As the object is raised to 45 m above the ground, 1.8 s is too short for the object fall 45 m under constant acceleration.

The resultant force is different for a rebounded case and non-rebounded case.

Assume that there is no rebound, the question did not specify the time of contact between the object and the ground, instead, it states “It takes 1.8 s for it to fall from rest to the ground.”

As I mentioned it is too short of time for the object to undergo free fall from 45 m above the ground. We can calculate the time of free fall by using

s = ½ at2

which would give time of free fall to be 3.0 s.

It seems that 1.8 s is the time of contact between the object and the ground rather than the time that the object falls from rest to the ground.
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2 years ago
#17
(Original post by jmushtaq)
We have an object that is raised 45m above the ground. It takes 1.8s for it to fall from rest to the ground. The mass of the object is 90kg. What is the resultant force by which the object hits the ground? Assume air resistance and friction is negligible.
If you rephrase the question, I would suggest that you post the original question.

If this is the original question, I would say it is a bad one. 0
2 years ago
#18
(Original post by Eimmanuel)
As I mention there are missing info or the phrasing of the question seems a bit odd and it seems that you refuse to accept it

I would elaborate further.
I assume that the object in on Earth surface.
As the object is raised to 45 m above the ground, 1.8 s is too short for the object fall 45 m under constant acceleration.

The resultant force is different for a rebounded case and non-rebounded case.

Assume that there is no rebound, the question did not specify the time of contact between the object and the ground, instead, it states “It takes 1.8 s for it to fall from rest to the ground.”

As I mentioned it is too short of time for the object to undergo free fall from 45 m above the ground. We can calculate the time of free fall by using

s = ½ at2

which would give time of free fall to be 3.0 s.

It seems that 1.8 s is the time of contact between the object and the ground rather than the time that the object falls from rest to the ground.
I agree, the wording is poor.

I'm pretty sure it's wanting you to calculate the force of gravity - not assume it's on earth hence why you're given the time.
Think it's wanting you to think it's a non rebound. - don't know if you learn about rebound at AS level. I'm doing higher RN so..

I mentioned the bit in bold in another post of mines, mentioned the acceleration must be 13.8888 (gravity = 13.8888) or that an external force was applied to the object from above. - though, if from rest then external force wouldn't be accounted for so there'd be a difference in gravity from earths.

The time of contact between ball and ground would be 1.8 seconds in, and with no rebound then it'd stay there for the rest of time.

I think at the end you're a tad confused maybe, if it was the time of contact between ball and ground then that'd be weird as the question's asking about the force of impact. (may just be reading what you're saying wrong tho)
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2 years ago
#19
(Original post by Relentas)
....

I'm pretty sure it's wanting you to calculate the force of gravity - not assume it's on earth hence why you're given the time.
...
I doubt so.

(Original post by Relentas)
....I think at the end you're a tad confused maybe, if it was the time of contact between ball and ground then that'd be weird as the question's asking about the force of impact. (may just be reading what you're saying wrong tho)
To calculate impact force, it needs time of contact or what is known as duration of impact.
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2 years ago
#20
(Original post by jmushtaq)
We have an object that is raised 45m above the ground. It takes 1.8s for it to fall from rest to the ground. The mass of the object is 90kg. What is the resultant force by which the object hits the ground? Assume air resistance and friction is negligible.
(Original post by Relentas)
I agree, the wording is poor.

I'm pretty sure it's wanting you to calculate the force of gravity - not assume it's on earth hence why you're given the time.
Think it's wanting you to think it's a non rebound. - don't know if you learn about rebound at AS level. I'm doing higher RN so..

I mentioned the bit in bold in another post of mines, mentioned the acceleration must be 13.8888 (gravity = 13.8888) or that an external force was applied to the object from above. - though, if from rest then external force wouldn't be accounted for so there'd be a difference in gravity from earths.

The time of contact between ball and ground would be 1.8 seconds in, and with no rebound then it'd stay there for the rest of time.

I think at the end you're a tad confused maybe, if it was the time of contact between ball and ground then that'd be weird as the question's asking about the force of impact. (may just be reading what you're saying wrong tho)
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