ANGLES

[Zebra]

When you guys are working out the angle of something landing.

what do you guys do?

use tan (x),sin(x),cos(x) ? use cos (x) = i or j or k / mag(i,j,k)?

[Asx4Life]

tan x man. But do we use velocity or position vector?

[Zebra]

tan x man. But do we use velocity or position vector?

we use v.

we can in fact use any trig i figured.... hmmm
i just need to make sure i don't confuse this with the angle formed with x,y,z axis.

[b^3]

I think this is from the 2007 exam yeh? I had to look at the itute solutions to do this but anyway. The velocity vector gives you what direction you are travelling in (and how fast), so it is the place to start.

Now what the question is actually asking is what angle (from the horizontal) is it landing at. i.e. How sharply is it coming in.



So to get that angle there we need to use
degrees. Hope that helps.

[b^3]

dam can't edit anymore, take theat extra component thing away and I missed the line

[Zebra]

thanks for that ! but remember, sin cos tan will get you the same answer if you know what you are doing! (i fink...)

[b^3]

thanks for that ! but remember, sin cos tan will get you the same answer if you know what you are doing! (i fink...)

Yeh as long as you have the right values in, and I think VCAA used sin for it but the tan way made more sense to me.

[Zebra]

thanks for that ! but remember, sin cos tan will get you the same answer if you know what you are doing! (i fink...)

Yeh as long as you have the right values in, and I think VCAA used sin for it but the tan way made more sense to me.

I see. hmmm i shall go to bed in about half an hour. GOOD LUCK man!

[Zebra]

wait, if you were to find the angle in mid air..... do you find the derivative and than

tan (x) = derivative?


i'm pretty sure.

[b^3]

wait, if you were to find the angle in mid air..... do you find the derivative and than

tan (x) = derivative?


i'm pretty sure.

Derivative, if you're talking the derivative of the position vector, which then gives the velocity vector then yes. Since this velocity is not time dependent, it is travelling in a straight line and so the angle is the same no matter what time it is so mid air works.

Also post number: 1234

[dc302]

If you're wondering how to remember whether to use position or velocity...

Velocity = speed and DIRECTION.

[funkyducky]

For the 2007 one, I took the magnitude of v at ground level (ie. sqrt of the i and j components squared) so i had a right angle triangle with the k-component as the height and then tan'd that. Wow this is an ambiguous explanation, but I cbf.

[b^3]

For the 2007 one, I took the magnitude of v at ground level (ie. sqrt of the i and j components squared) so i had a right angle triangle with the k-component as the height and then tan'd that. Wow this is an ambiguous explanation, but I cbf.

Yeh thats basically the same as I did above, I just didn't say that (I probably should have, it would have explained it better).

[jane1234]

I just used the cos(x) = c/|u| rule where u = ai + bj + ck and x is the angle it makes with the z axis, this gives an obtuse angle so just take 90 degrees from it and then you get the answer...

[jane1234]

While we are on the subject, I don't get 4d.

Why would the closest point to the base of the tower be when r.v = 0 ??