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Maths mechanics

I’ll post the question and my working out but idk how to get further
Reply 1
64C2CA6B-CBC3-4DDA-9246-7E206A29D2EF.jpegimage.jpgIt’s question 5
Reply 2
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.
(edited 2 years ago)
Reply 3
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
(edited 2 years ago)
Reply 4
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.
Reply 5
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
Reply 6
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.
(edited 2 years ago)
Reply 7
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
Reply 8
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 ...
(edited 2 years ago)
Reply 9
I got 3/5=gt^2
I need to either sub something into t or use another equation
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.
Reply 11
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.

image.jpg I got it
Original post by Heixi
image.jpg 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?
(edited 2 years ago)
Reply 13
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
Original post by Heixi
Oh yeah

It should just be a couple of lines, hence the 2 marks for the question part.
(edited 2 years ago)
Reply 15
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
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).
Reply 17
Something like
image.jpg
Probably, but what do those equations represent?
Why can you sub time from one into the other?

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