# Isaac Physics Flying Return

https://isaacphysics.org/questions/flying_return?board=cbc6660d-5de6-4207-8783-5be3d57c87d1&stage=a_level

I've been stuck on this question for a while and I can't see what I'm doing wrong.

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Original post by mosaurlodon
https://isaacphysics.org/questions/flying_return?board=cbc6660d-5de6-4207-8783-5be3d57c87d1&stage=a_level

I've been stuck on this question for a while and I can't see what I'm doing wrong.

What did you try?

This is the working I've done so far
Original post by mosaurlodon

This is the working I've done so far

As the diagram in hint 3 illustrates, v will not be directed along the from A to B, it must be at an angle to the path to counteract the head/side wind.

Just working it through now.
Original post by mosaurlodon

This is the working I've done so far

Getting isaac to recognise cos as cos, rather than 3 variables ... grrr.

Anyway, think about a velocity isosceles triangle as per hint 3. The base angles are alpha and you can get sin(alpha) in terms of k and sin(theta) using the sin rule. Then project onto the base (direction of motion) to get the velocities as
vcos(alpha) +/-kvcos(theta)
then get the difference in times as youve tried to do.
If you have an equation for sin(alpha), how would you find cos(alpha)?

so cos(alpha) would be cos(arcsin(k*sin(theta))) -> is there a rule for this I am unsure how to simplify
Original post by mosaurlodon
If you have an equation for sin(alpha), how would you find cos(alpha)?

so cos(alpha) would be cos(arcsin(k*sin(theta))) -> is there a rule for this I am unsure how to simplify

Its just pythagoras so
sin^2 + cos^2 = 1
so
cos = ...

Just think of a unit (hypotenuse) right triangle with base cos and height sin and everything falls out of that.
so the time difference should be x/v - x/2vkcos(theta) right?
Original post by mosaurlodon
so the time difference should be x/v - x/2vkcos(theta) right?

No, youd have (with appropriate substitution)
v1,v2 = vcos(alpha) +/- kvcos(theta)
so
x/v1 - x/v2
and simplify it. If thats what youve tried to do, post your working.
(edited 2 months ago)
sorry to have this take so long, but are my velocities correct?

so Vab = vcos(alpha) + kvcos(theta) = v(2kcos(theta))
and Vba = vcos(alpha) - kvcos(theta) = v
-> from the equation sin(alpha) = k*sin(theta)
so cos(alpha) = sqrt(1-k^2sin^2(theta)) = kcos(theta)
Original post by mosaurlodon
sorry to have this take so long, but are my velocities correct?

so Vab = vcos(alpha) + kvcos(theta) = v(2kcos(theta))
and Vba = vcos(alpha) - kvcos(theta) = v

No, alpha and theta are not the same angle. The headwind is at an angle theta, the driving velocity is at an angle alpha, both relative to the resultant direction of motion.

You "know" cos(theata), but have to get an expression for cos(alpha) in terms of sin(theta) as per the previous posts.
Original post by mosaurlodon
-> from the equation sin(alpha) = k*sin(theta)
so cos(alpha) = sqrt(1-k^2sin^2(theta)) = kcos(theta)

cos(alpha) = sqrt(1-k^2sin^2(theta))
is correct. Its not equal to kcos(theta).
(edited 2 months ago)
Oh my gosh, I can't believe it my answer was right

Thank you so much!!
my mistake was k^2sin^2(theta)+k^2cos^2(theta)=1 rather than k^2sin^2(theta)+k^2cos^2(theta)=k^2
ok I was working on part b of this question, and my monkey brain decided to randomly replace theta with phi, and it somehow turned out to be the right answer - can someone explain this to me?
Original post by mosaurlodon
ok I was working on part b of this question, and my monkey brain decided to randomly replace theta with phi, and it somehow turned out to be the right answer - can someone explain this to me?

Can you post what you did?

I also am confused on which way the wind is blowing, in the q, it says from north to south, but in the hint 2, it says from west to east.
Original post by mosaurlodon

I also am confused on which way the wind is blowing, in the q, it says from north to south, but in the hint 2, it says from west to east.

Ok, just wanted to make sure. The hint 2 part is wrong, and really if you rotated the diagram (hint 3) round from part a) so that the wind was blowing northerly and phi is the bearing (relative to north), then its just theta from part a).

Most of the question part info is irrelevant, you simply need the direction of the headwind relative to your direction of motion (as you should expect).

is the following diagram a good illustration of what you're saying?