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Rafy

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Raolan

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Somehow for (b)iii. I managed to get 15 <= x <= ~23, but also x >= ~50 (x = feta). So did two friends.

Anyone get the actual answer?

The geometry question was really easy, though I'm assuming there was a simpler method for the first part rather than proving all right angles by angle in a semicircle = 90 degrees.

Burned out on the last one though. Was low on time and kind of just fumbling around, so I went back and finished the Newton's Method in the last 2 minutes rather than bothering with it.
 

kliro_10

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Somehow for (b)iii. I managed to get 15 <= x <= ~23, but also x >= ~50 (x = feta). So did two friends.

Anyone get the actual answer?
yes, you have to consider the other interval of sin2(theta) so you end up with

theta>50

and

67< theta < 75
 

RishBonjour

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yes, you have to consider the other interval of sin2(theta) so you end up with

theta>50

and

67< theta < 75
OMGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG.
FMLLLLL. NO WONDER.

I got 15<x<23
 

TheOptimist

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Somehow for (b)iii. I managed to get 15 <= x <= ~23, but also x >= ~50 (x = feta). So did two friends.

Anyone get the actual answer?

The geometry question was really easy, though I'm assuming there was a simpler method for the first part rather than proving all right angles by angle in a semicircle = 90 degrees.

Burned out on the last one though. Was low on time and kind of just fumbling around, so I went back and finished the Newton's Method in the last 2 minutes rather than bothering with it.
Thats what i got, not sure if its right though...
 

GoldyOrNugget

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66.95<= @ <= 75. Solving for x inside the range gave you two solutions because sin(@) = sin(180-@). Taking the intersection of these regions with the range provided by the y-limitation (x >= 50 or something) gave you the above solution.

For bi), use cos rule and Pythagoras to solve for r. You'll end up with a sin(@) that will cancel in the root as u^2(sin^2(@) + cos^2(@)) = u^2.

For part bii), t=5 min is 1/12 hours. Solve dr/dt = dr/du * du/dt where du/dt = 360km/h. This will give you an expression in terms of u. Now at t=5, the plane has flown 1/12 * 360km = 30km. So sub in u=30 and you get your answer :)
 

Nooblet94

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For b)iii) there's two possible intervals theta can lie in in order to satisfy the distance requirement, but only one satisfies the height.
 

HeroicPandas

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yes, you have to consider the other interval of sin2(theta) so you end up with

theta>50

and

67< theta < 75
You are correct, mostly everyone got up to the 2 sets of values, then the hard part was to add the inequalities

OR the best way was check if the values satisfy both equations (x and y) then you will see that u must add the 2 inequalities :D

good work kliro_10
 

GoldyOrNugget

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Nooblet94 are you sure there are 2 cos(alphas)? If you differentiate dr/du via chain rule, the cos(alpha) inside the brackets will differentiate to a -sin(alpha) outside.
 

Nooblet94

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Nooblet94 are you sure there are 2 cos(alphas)? If you differentiate dr/du via chain rule, the cos(alpha) inside the brackets will differentiate to a -sin(alpha) outside.
alpha's a constant, so cos(alpha) is also a constant.
 

Nooblet94

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My upper bound for the projectile question is incorrect.
 

osak23

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for b iii)
were you just meant to sub (125,150) into the parabloa equation to get the thetas..

........
 

nofate

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Do you reckon it will matter if I did sqroot(900 + 1 - 30cosa) instead of sqroot(901 - 30cosa)?
 

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