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# M1 suvat problem watch

1. Hey, just started M1 mechanics and to be honest I'm finding it pretty difficult right now. I'm stuck on this question and have no idea on how to get to an answer.

The question:

A body falls from the top of a tower. During the last second of its motion is falls (7/16) of the whole distance. Show that the time taken for the descent is independent of the value g, and find the height of the tower in terms of g.

The answers: time 4s and height 8g metres

I have no idea how to do this so I would be very grateful for an explanation!
2. (Original post by Wright98)
Hey, just started M1 mechanics and to be honest I'm finding it pretty difficult right now. I'm stuck on this question and have no idea on how to get to an answer.

The question:

A body falls from the top of a tower. During the last second of its motion is falls (7/16) of the whole distance. Show that the time taken for the descent is independent of the value g, and find the height of the tower in terms of g.

The answers: time 4s and height 8g metres

I have no idea how to do this so I would be very grateful for an explanation!
Always best to post in the maths subforum rather than this umbrella one.

That said.

The body falls, rather than is projected, so initial velocity is 0

Our equation of motion is therefore s= 1/2 (at^2)

Let's say it falls for a total time of t.
So, we have one equation s= 1/2 (at^2)

So, prior to the final second it will have been falling for t-1 seconds, and gone a distance s' (as yet unknown)m and we have a second equation s' = 1/2 (a(t-1)^2)

We now need to relate these to and get rid of the s and s'

We know it falls 7/16 of the distance in the last second, so if falls 9/16 of the distance prior to that.

So, s'/s = 9/16

Plug in your values and solve for t.

PS: Quote if you need a reply as I don't often visit this forum.
3. (Original post by ghostwalker)
Always best to post in the maths subforum rather than this umbrella one.

That said.

The body falls, rather than is projected, so initial velocity is 0

Our equation of motion is therefore s= 1/2 (at^2)

Let's say it falls for a total time of t.
So, we have one equation s= 1/2 (at^2)

So, prior to the final second it will have been falling for t-1 seconds, and gone a distance s' (as yet unknown)m and we have a second equation s' = 1/2 (a(t-1)^2)

We now need to relate these to and get rid of the s and s'

We know it falls 7/16 of the distance in the last second, so if falls 9/16 of the distance prior to that.

So, s'/s = 9/16

Plug in your values and solve for t.

PS: Quote if you need a reply as I don't often visit this forum.
First of all thanks for the help However, when I add in the values of 9 and 16 for the distance one second before reaching the ground and the total distance, I am still unable to work out what the time is due to acceleration having to be in the form of g. If I were to leave in acceleration as g, then how is it possible to reach a time of 4 seconds?
4. (Original post by Wright98)
First of all thanks for the help However, when I add in the values of 9 and 16 for the distance one second before reaching the ground and the total distance, I am still unable to work out what the time is due to acceleration having to be in the form of g. If I were to leave in acceleration as g, then how is it possible to reach a time of 4 seconds?
The g's cancel, and you can rearrange to get a quadratic in t.

If it's not coming out, post some working and we can see what's going on.
5. (Original post by ghostwalker)
The g's cancel, and you can rearrange to get a quadratic in t.

If it's not coming out, post some working and we can see what's going on.

Okay, I have an answer of 4 but also 0.6 seconds. Why is 4 the only correct answer? Also, is my method correct because it seems quite a long approach for a question which first seemed quite simple? Mind you, it probably is simple I'm just finding this pretty difficult at the moment :/
6. (Original post by Wright98)

Okay, I have an answer of 4 but also 0.6 seconds. Why is 4 the only correct answer?
Given that it fell 7/16 of the distance in the final second, t must be greater than 1.

Also, is my method correct
Huh! I presume you're using the method I outlined.

because it seems quite a long approach for a question which first seemed quite simple? Mind you, it probably is simple I'm just finding this pretty difficult at the moment :/
This type of question does require more thought, and that's where the time comes in. The shortness of the question is deceptive - can't think of a quicker way to do it.
7. (Original post by ghostwalker)
Given that it fell 7/16 of the distance in the final second, t must be greater than 1.

Huh! I presume you're using the method I outlined.

This type of question does require more thought, and that's where the time comes in. The shortness of the question is deceptive - can't think of a quicker way to do it.
Okay, thank you very much for your help! I am aware that I've asked a lot questions, sorry about that but as I said I'm finding it difficult at the moment
8. (Original post by Wright98)
Okay, thank you very much for your help! I am aware that I've asked a lot questions, sorry about that but as I said I'm finding it difficult at the moment
Asking questions is not a problem.

But do use the maths subforum, it's a lot more active than this umbrella one.
9. (Original post by Wright98)
...
Sorry, just saw your working (missed it previously - another reason to use the maths forum - plenty of people to check responses)

Yes - it's fine.

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