Heixi
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#1
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#1
I’ll post the question and my working out but idk how to get further
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Heixi
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#2
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#2
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Size:  168.1 KBIt’s question 5
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mqb2766
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#3
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#3
Youve over complicated it. Just think about the answer.
What suvat equation(s) can you use based on the property about the maximum vertical height?
Also make sure you get the displacement correct.
Last edited by mqb2766; 4 days ago
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Heixi
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#4
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#4
(Original post by mqb2766)
Youve over complicated it. Just think about the answer.
What suvat equation(s) can you use based on the property about the maximum vertical height?
Also make sure you get the displacement correct.
Then is the displacement just 2.8m?
Would I use s=Ut +1/2at^2 rearranged for U ?
I’ve got my horizontal and vertical the wrong way round
Last edited by Heixi; 4 days ago
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mqb2766
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#5
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#5
(Original post by Heixi)
Then is the displacement just 2.8m?
Would I use s=Ut +1/2at^2 rearranged for U ?
There is no reference to time in the question part, so no.
The ball reaches a height of 2.8 above the ground. So the displacement is ....
You've drawn a diagram which is good, but you've incorrectly marked on the maximum height.
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Heixi
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#6
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#6
(Original post by mqb2766)
There is no reference to time in the question part, so no.
The ball reaches a height of 2.8 above the ground. So the displacement is ....
You've drawn a diagram which is good, but you've incorrectly marked on the maximum height.
Then 2.8-2.5=0.3m?
It says above the ground so I assumed 2.8m above the point
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mqb2766
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#7
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#7
(Original post by Heixi)
Then 2.8-2.5=0.3m?
It says above the ground so I assumed 2.8m above the point
Yes. The initial point is 2.5 above the ground, so the vertical dispacement from the initial point is 0.3.
Last edited by mqb2766; 4 days ago
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Heixi
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#8
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#8
(Original post by mqb2766)
Yes. The initial point is 2.5 above the ground, so the vertical dispacement from the initial point is 0.3.
So I would use v^2=U^2+2as vertically
but isn’t U at max height =0
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mqb2766
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#9
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#9
(Original post by Heixi)
So I would use v^2=U^2+2as vertically
but isn’t U at max height =0
No, U = ... is what you have to show.
However, you've almost got the right idea ...
Last edited by mqb2766; 4 days ago
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Heixi
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#10
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#10
I got 3/5=gt^2
I need to either sub something into t or use another equation
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mqb2766
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#11
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#11
(Original post by Heixi)
I got 3/5=gt^2
I need to either sub something into t or use another equation
Its not that suvat equation as time is not involved in this question part.
Your previous post had the right suvat equation to use for vertical motion.
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Heixi
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#12
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#12
(Original post by mqb2766)
Its not that suvat equation as time is not involved in this question part.
Your previous post had the right suvat equation to use for vertical motion.
Name:  image.jpg
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mqb2766
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#13
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#13
(Original post by Heixi)
Name:  image.jpg
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Size:  124.2 KB I got it
You;ve got a small sign error at the start, just think about the direction of a and s. Also, your square gets dropped, then reappears?
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Heixi
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#14
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#14
(Original post by mqb2766)
You;ve got a small sign error at the start, just think about the direction of a and s. Also, your square gets dropped, then reappears?
Oh yeah
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mqb2766
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#15
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#15
(Original post by Heixi)
Oh yeah
It should just be a couple of lines, hence the 2 marks for the question part.
Last edited by mqb2766; 4 days ago
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Heixi
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#16
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#16
(Original post by mqb2766)
It should just be a couple of lines, hence the 2 marks for the question part.
I’ve sorted it , how should I go about b ? I’ve done the rest
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mqb2766
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#17
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#17
(Original post by Heixi)
I’ve sorted it , how should I go about b ? I’ve done the rest
Based on your diagram and the form of the answer, do you have any ideas?
From part a) you have the initial speed (as a function of alpha).
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Heixi
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#18
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#18
Something like
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mqb2766
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#19
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#19
Probably, but what do those equations represent?
Why can you sub time from one into the other?
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