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Mechanics

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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Reply 2
Avatar for mqb2766
mqb2766
21
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?
(edited 3 years ago)
CF25B2C8-C7CE-4825-87F9-88B0EAA9FB0F.jpegI’ve got a quadratic as expected but there are two unknown values?
Reply 4
Avatar for mqb2766
mqb2766
21
Original post by M.Johnson2111
CF25B2C8-C7CE-4825-87F9-88B0EAA9FB0F.jpegI’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.
(edited 3 years ago)
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?
Reply 6
Avatar for mqb2766
mqb2766
21
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.
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?
Reply 8
Avatar for mqb2766
mqb2766
21
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? :banghead:
(edited 3 years ago)
Sorry Im looking at the wrong question

3u^2/2g? for displacement
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.
Is this for 3c?
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?
Original post by M.Johnson2111
Is this for 3c?

That's the question Im answering.
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)?
Okay so from what I know T = u/g and H = u^2/2g
S = ut + 1/2at^2
Ive literally just got my axis labelled, trying to find the H and T values to be able to sketch
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)
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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