yea same having trouble with the second part
I thought it would be nice to have a thread for answering past HSC questions and any other for that matter, in the lead up to the exam in a few weeks. So here it is.
I'll start off with one from the 2001 HSC which I have been struggling with. Any help would be appreciated.
http://prntscr.com/kumrq3
yea same having trouble with the second part
Consider .
By the cosine rule,
Equating and dividing out by (the height cannot be zero),
Last edited by fan96; 15 Sep 2018 at 6:10 PM.
Applying the auxiliary angle transformation:
* Note the restriction in the diagram - we do not need to take a general solution here.
.
(Since )
Since this is the only turning point within the given domain and , this is clearly a minimum turning point.
If we put on the vertical axis, the graph of this function will be similar in shape to a concave up parabola.
Note when sketching that are actually excluded from the domain, so you should draw an open circle at the endpoints of the domain to indicate that they are not part of the function.
Last edited by fan96; 15 Sep 2018 at 9:43 PM.
http://prntscr.com/kvbms7
Thanks
Last edited by integral95; 17 Sep 2018 at 9:40 PM.
“Smart people learn from their mistakes. But the real sharp ones learn from the mistakes of others.”
― Brandon Mull
http://prntscr.com/kw5d3q
Part iii. Thanks
The horizontal range of the ball is maximised when . This can be easily shown by examining the equation in ii).
Case 1: The ceiling allows for an angle of projection of .
If the ceiling allows for at least this angle of projection, then is the optimal angle to throw the ball.
It doesn't matter how tall the ceiling is after a certain point - because the best angle is and any higher angle will simply lower .
The horizontal range obtained by projecting at this optimal angle is then .
But when does the ceiling allow for this angle? It is when the maximum height of the projectile (with angle of projection ) is less than or equal to the height allowed by the ceiling (which is ).
Mathematically, this can be represented by the condition
.
Case 2: The ceiling does NOT allow for an angle of projection of .
As decreases past , so too does , from ii).
Therefore we want the largest value of we can get. This value occurs when the tip of the ball's trajectory coincides with the ceiling, i.e.
Rearrangement gives:
Finally, to get the answer, substitute (1) and (2) into:
Last edited by fan96; 19 Sep 2018 at 10:09 PM.
Could I get some help with part i and ii please. Thanks
http://prntscr.com/kz7x4d
1-on-1 Maths Tutoring(IB & HSC): Epping, Beecroft, Eastwood, Carlingford & Beyond
IB: Maths Studies, Maths SL & Maths HL; HSC: 2U, 3U & 4U
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i) Hint: the expression you need to prove can be rearranged to
ii) Starting from the inductive hypothesis:
From i) by rearranging we get .
This seems to be my thread haha.
If anyone could help with this, and maybe some tips on how to approach these sort of problems- I'm not great at them.
A die is loaded in such a way that in 8 throws of the die, the probability of getting 3 even numbers is four times the probability of getting 2 even numbers. Find the probability that a single throw of the die results in an even number.
This is a Binomial Probability question.
Solving you get the P(even) in a single throw = p = 2/3.
Last edited by Drongoski; 7 Oct 2018 at 8:17 AM.
1-on-1 Maths Tutoring(IB & HSC): Epping, Beecroft, Eastwood, Carlingford & Beyond
IB: Maths Studies, Maths SL & Maths HL; HSC: 2U, 3U & 4U
Highly Qualified & Highly Experienced. Estimated ATAR > 9.995
There are IB Maths Tutors and there are IB Maths Tutors.
May I ask where you found that proj motion question?
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