# Mechanics

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#1
Hi,
Please can I have some help with part c of this question?

For part a I got, T = u/g

For part b I got, H = u^2/2g

So far I have found 3H = 3u^2/2g and using SUVAT

S = 3u^2/2g
U = u
V = ?
A = g
T = ? X - trying to solve

And then I was using my answer for part a , T and going to add this to the value of T when falling?

Any help please as cant get past this point yet?
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#2
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1 month ago
#3
(Original post by M.Johnson2111)
Hi,
Please can I have some help with part c of this question?

For part a I got, T = u/g

For part b I got, H = u^2/2g

So far I have found 3H = 3u^2/2g and using SUVAT

S = 3u^2/2g
U = u
V = ?
A = g
T = ? X - trying to solve

And then I was using my answer for part a , T and going to add this to the value of T when falling?

Any help please as cant get past this point yet?
You know s,a,u and want t. Which suvat is it?
Or think about the s-t relationship, how long does it take to fall double, quadruple the distance in terms of T? Why 3H, how far from the peak is that?
Last edited by mqb2766; 1 month ago
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#4
I’ve got a quadratic as expected but there are two unknown values?
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1 month ago
#5
(Original post by M.Johnson2111)
I’ve got a quadratic as expected but there are two unknown values?
Sure, I was expecting you to reject the suvat approach for this reason (messy to sort out). Sketch the quadratic (height) trajectory against time. What points can you mark on.
Last edited by mqb2766; 1 month ago
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#6
How would I do this rejecting the SUVAT approach?

So to sketch this would I let ut = x or t = x? from my equation

If I let ut = x, there is a repeated root of x = 9.8
Would this be my final answer, t = 9.8? I dont quite understand how thatd make sense but by substitution I think mathematically accepted?
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1 month ago
#7
a) and b) strongly suggest splitting up the trajectory into time to peak, peak to floor. What are the displacements and times "travelled" during those times. Remember displacement is a quadratic function of time.
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#8
Okay so a graph for time to peak and one for time to floor?
S when peaking is 885
Then S from peak to floor = -885?
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1 month ago
#9
(Original post by M.Johnson2111)
Okay so a graph for time to peak and one for time to floor?
S when peaking is 885
Then S from peak to floor = -885?
885?
Last edited by mqb2766; 1 month ago
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#10
Sorry Im looking at the wrong question

3u^2/2g? for displacement
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1 month ago
#11
(Original post by M.Johnson2111)
Sorry Im looking at the wrong question

3u^2/2g? for displacement
Have you sketched the quadratic displacement-time curve? If so, can you upload it.
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#12
Is this for 3c?
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#13
Right so on the y-axis, I have H which is the displacement and then time on the x-axis

Where does the quadratic, equation involving T come into this and do I use this?
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1 month ago
#14
(Original post by M.Johnson2111)
Is this for 3c?
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1 month ago
#15
(Original post by M.Johnson2111)
Right so on the y-axis, I have H which is the displacement and then time on the x-axis

Where does the quadratic, equation involving T come into this and do I use this?
Can you upload what you've sketched? Which suvat describes s-t relationship (not involving v)?
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#16
Okay so from what I know T = u/g and H = u^2/2g
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#17
S = ut + 1/2at^2
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#18
Ive literally just got my axis labelled, trying to find the H and T values to be able to sketch
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1 month ago
#19
(Original post by M.Johnson2111)
S = ut + 1/2at^2
So s-t is a quadratic curve with a max at the point in a) b)
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1 month ago
#20
(Original post by M.Johnson2111)
Ive literally just got my axis labelled, trying to find the H and T values to be able to sketch
A sketch doesn't have to be exact. You know the relative heights and times, that's all that's needed.
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