Harder 3U questions (1 Viewer)

[apollo1]

need help with last two parts of this question:





last three parts of this question:

 

[b3kh1t]

With the circle geometry question, for ii;

() is a fixed angle as it lies on the fixed chord BC.
Therefore in triangle AEC, and are constants, thus is a constant angle.
Since the constant angle () subtends the chord ED on the circle BCDE, then ED is a constant length.

for iii;

Let M be the midpoint of ED and L the midpoint of BC, and consturct ML and EL.
EM=MD ( M is the midpoint of ED)
therefore EM=1/2ED ( a constant value)
Also EL is a constant value as it is the radius of the circle BCDE.
ML is perpendicular to ED ( perpendicular line from the centre of the circle BCDE bisects the chord ED)



(a constant value, as EL and EM are constants)

Since ML is a constant, it is the radius of the locus M which lies on the circle with the with centre L.

 

[Drongoski]

b3kh1t

Nice proof.

 

[math man]

For the projectile question, your eqns of motion for the slow particle should be:



For the faster one:



Now we want to find t when y=0 for the slower particle, which gives:



now we are told at this point the faster particle makes an angle beta downwards, so if you we look at the question we need to get a tan beta in there and this is done by finding the velocity at that specific point for the faster particle:

note it is important we have the neg sign for y' as we are travelling downwards. When deriving the eqns of motion for the faster particle you should of got:



subbing these into our equation for tan beta and subbing our t value at this point gives:



For ii) sub:



which gives:



Simplying this gives:



We simply this to :



as: