Can anyone give me a few hints for 12b. and 6?
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pippabethan
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- 12-03-2016 16:03
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- 12-03-2016 16:28
(Original post by pippabethan)
Can anyone give me a few hints for 12b. and 6?
Horizontally:(x is horizontal displacement)
Vertically:(y is vertical displacement)
Path of the projectiles means an equation of the form, so you want to eliminate
from the second equation.
You know that, so substitute that into your second equation and that'll give you the path (which will be a quadratic function of
, plotting it will look familiar to you, it'll be a maximum quadratic that models the path of the projectile quite literally).
Last edited by Zacken; 13-03-2016 at 09:39. -
thebrahmabull
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- 12-03-2016 18:08
(Original post by pippabethan)
Can anyone give me a few hints for 12b. and 6?
For part 12B the idea is that they have different constant horizontal speeds I.e. rate of change of distance with time is different for each. I would like to see the solutions if you do them! -
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- 12-03-2016 18:20
(Original post by thebrahmabull)
12 looks badass. I haven't seen this type before. For part 12a I think the idea is gradient=acceleration =g. But how did you mathematically show it? Mind sharing?
For part 12B the idea is that they have different constant horizontal speeds I.e. rate of change of distance with time is different for each. I would like to see the solutions if you do them! -
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- 12-03-2016 18:50
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- 12-03-2016 19:44
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- 12-03-2016 19:57
(Original post by TeeEm)
we are waiting ... -
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- 12-03-2016 19:59
(Original post by Zacken)
As I am for the OP to reply...
I thought you were putting them up. -
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- 12-03-2016 20:47
(Original post by pippabethan)
Can anyone give me a few hints for 12b. and 6?
So
The gradient of this line should be fairly easy to find, and it will be in terms of onlyand
, which are constants.
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pippabethan
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- 12-03-2016 22:00
(Original post by Zacken)
For Q6, resolve the velocity horizontally and vertically.
Horizontally:(x is horizontal displacement)
Vertically:(y is vertical displacement)
Path of the projectiles means an equation of the form, so you want to eliminate
from the second equation.
You know that, so substitute that into your second equation and that'll give you the path (which will be a quadratic function of
, plotting it will look familiar to you, it'll be a maximum quadratic that models the path of the projectile quite literally).
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pippabethan
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- 12-03-2016 22:02
Sorry everyone, my wifi stopped working all day :/ Thank you so much for your help!! I can post the answers now thebrahmabull if you still want them?
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pippabethan
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- 12-03-2016 22:05
(Original post by Zacken)
Oh well, since I've done Q12 - might as well put this up, since you're not replying:
So
The gradient of this line should be fairly easy to find, and it will be in terms of onlyand
, which are constants.
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- 12-03-2016 22:08
(Original post by pippabethan)
I got the constant gradient part, but how do you find the rate of increase of distance? I think the formula v^2=u^2 + 2as is used because the answer involves a root. -
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- 12-03-2016 22:46
(Original post by Zacken)
Increasing at a constant rate means that the gradient of the distance separating them is positive. Can you prove that your answer to part (a) is positive? -
pippabethan
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- 13-03-2016 09:36
(Original post by Zacken)
For Q6, resolve the velocity horizontally and vertically.
Horizontally:(x is horizontal displacement)
Vertically:(y is vertical displacement)
Path of the projectiles means an equation of the form, so you want to eliminate
from the second equation.
You know that, so substitute that into your second equation and that'll give you the path (which will be a quadratic function of
, plotting it will look familiar to you, it'll be a maximum quadratic that models the path of the projectile quite literally).
-
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- 13-03-2016 09:41
(Original post by pippabethan)
Wouldn't t=x/20 ? I got 400y = 600x -4.9x^2 as the final answer, but the book says it's 900ydirection would be horizontal, so the initial horizontal speed is
, so it would be
.
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pippabethan
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- 13-03-2016 10:04
I don't see how the rate of increase should equal u(2(1-cos(theta - alpha))^0.5.
For a) I got (sin theta - sin alpha)/(cos theta - cos alpha) -
thebrahmabull
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- 13-03-2016 11:00
(Original post by Zacken)
Increasing at a constant rate means that the gradient of the distance separating them is positive. Can you prove that your answer to part (a) is positive? -
thebrahmabull
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- 13-03-2016 11:17
(Original post by TeeEm)
we are waiting ...
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