davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
64/81 implies a a 0.7901 probability of winning
64/81 implies a a 0.7901 probability of winning
HSC 2015 MX1 Marathon (archive) (1 Viewer)
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davidgoes4wce
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Re: HSC 2015 3U Marathon
0.7901 in betting odds is around $1.26 to win a match (In terms of TAB or Betfair odds)
it makes sense because if Novak is 2/3 ($1.50 to win a set), usually the better player will win the match over longer sets.
0.7901 in betting odds is around $1.26 to win a match (In terms of TAB or Betfair odds)
it makes sense because if Novak is 2/3 ($1.50 to win a set), usually the better player will win the match over longer sets.
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
Each Set in TAB or Betfair odds would be
Novak $1.50 v $3.00 Andy
based on the probabilities assigned.
Each Set in TAB or Betfair odds would be
Novak $1.50 v $3.00 Andy
based on the probabilities assigned.
trumanblack
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Re: HSC 2015 3U Marathon
Q. A person walking alonga straight road observes a tower bearing 045degreesT, the angle of elevation being 6degrees. After traveling a distance of 5000m, the tower beares 315degreesT and the angle of elevation is 8degrees.
a) Find the height of the tower (to 0.1m)
b) Determine the angle which the road makes with a line bearing 090degreesT
Part a is okay but I can't visualise part b
For reference, question is from BHHS 1996 trial paper - but the solution to b) is messy and slightly cut off
Q. A person walking alonga straight road observes a tower bearing 045degreesT, the angle of elevation being 6degrees. After traveling a distance of 5000m, the tower beares 315degreesT and the angle of elevation is 8degrees.
a) Find the height of the tower (to 0.1m)
b) Determine the angle which the road makes with a line bearing 090degreesT
Part a is okay but I can't visualise part b
For reference, question is from BHHS 1996 trial paper - but the solution to b) is messy and slightly cut off
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
I got stumped by this projectile motion question tonight. Not going to look at the solution as I find discussing it with others more learner friendly.
I got stumped by this projectile motion question tonight. Not going to look at the solution as I find discussing it with others more learner friendly.
Re: HSC 2015 3U Marathon
If I recall correctly, the earlier parts help.I got stumped by this projectile motion question tonight. Not going to look at the solution as I find discussing it with others more learner friendly.
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
This was my effort:
This was my effort:
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
my thinking is halfway the horizontal distance is d/2 and using that information we can find alpha value
my thinking is halfway the horizontal distance is d/2 and using that information we can find alpha value
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
thinking of drawing a triangle and finding out the hypotenuse length of that, so we can eliminate 'cos a'
thinking of drawing a triangle and finding out the hypotenuse length of that, so we can eliminate 'cos a'
Re: HSC 2015 3U Marathon
 and the horizontal (I assume, unless $\alpha$ happens to be from a previous part and was defined differently, but I don't think it was). The correction statement would be that $\tan \beta=\frac{H-S}{\frac{d}{2}}$, where $\beta$ is the angle between the \emph{chord} (line segment) between $\mathcal{S}$ and $\mathcal{M}$ and the horizontal, where $\mathcal{M}$ is the maximum point (vertex) of the parabola. From a diagram, you can easily see that $\alpha>\beta$, and $\alpha$ and $\beta$ are both acute, so $\tan \alpha > \tan \beta=\frac{H-S}{\frac{d}{2}}$.)
my thinking is halfway the horizontal distance is d/2 and using that information we can find alpha value
Last edited:
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
OK I admit I stuffed that one up......I should have known the angle gets smaller as it leaves from the shoulders
OK I admit I stuffed that one up......I should have known the angle gets smaller as it leaves from the shoulders
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
It's a 4 mark question but I am still thinking about it
It's a 4 mark question but I am still thinking about it
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
ok after almost 25 minutes I think I have got the working out
ok after almost 25 minutes I think I have got the working out
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
wait false alarm, I'm close to the answer but not quite right
wait false alarm, I'm close to the answer but not quite right
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
Would x=d , when y=0?
Would x=d , when y=0?
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
I might have to get a personal tutor, still not good enough at cracking the last questions..........and in terms of time , I'm not answering these questions efficiently enough
I might have to get a personal tutor, still not good enough at cracking the last questions..........and in terms of time , I'm not answering these questions efficiently enough
Re: HSC 2015 3U Marathon
What are you using as y? Are you using it as y = 0 at the point where the ball was thrown? If so, the answer is yes.
Re: HSC 2015 3U Marathon
5C3 × (2/3)³ × (1/3)²
+ 5C4 × (2/3)^4 × (1/3)^1
+ 5C5 × (2/3)^5 × (1/3)^0
A more symmetrical calculation:part (ii)
I have uploaded my screenshot. Feel free if anyone sees any error in my work to correct me
5C3 × (2/3)³ × (1/3)²
+ 5C4 × (2/3)^4 × (1/3)^1
+ 5C5 × (2/3)^5 × (1/3)^0
davidgoes4wce
Well-Known Member
Re: HSC 2015 3U Marathon
I just had another attempt at that question 7b (iii) this morning, someone tell me where I am going wrong, I guess I'm still a bit confused about the angles used. Going to wave the white flag towards this question, if anyone has more helpful advice show it.
I just had another attempt at that question 7b (iii) this morning, someone tell me where I am going wrong, I guess I'm still a bit confused about the angles used. Going to wave the white flag towards this question, if anyone has more helpful advice show it.
Re: HSC 2015 3U Marathon
I just had another attempt at that question 7b (iii) this morning, someone tell me where I am going wrong, I guess I'm still a bit confused about the angles used. Going to wave the white flag towards this question, if anyone has more helpful advice show it.
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