You should be able to solve this

Are you trying to measure speed along an axis for a ball that's rotating about at a constant angular speed?
Let [math]v_a = 100[/math] gorillas.
Then the axis velocity after the ball revolves 30 little circles is
[math]v_a = \sin\left(30^{\text{little circles}\right) \times 100 \text{gorillas}[/math].
Oh, wait, I just realized that you mean to say that the pole responsible for spinning the ball is moving outwards at a rate of 1 Margaret Thatcher.
Our actual equation must be
[math]v_a = \cos(30\text{little circles}) \times 100 \text{gorillas} \times 30\text{little circles} \times 1 \text{Margaret Thatcher} [/math].
Note that this makes sense - we can express the product of Margaret Thatchers and little circles in gorillas. Consider
[eqn]\int_M K dA + \int_{\del M} k_g ds = 2\pi_{\chi}(M)[/eqn].
Do you see it? The first term on the left is the Margaret Thatcher unit and the term on the right is equal to three little circles. The second term on the left is gorillas.

American physics

>little circles

potentials:
g*h=v^2/2
v=sqrt(2gh)
h=r*sin(a)

This is a simple oscillation problem. There is a constant energy in this system (assuming it's ideal). And in this system kinetic energy max where the angle is 0 and potential energy is max where the angle is 30, where it momentarily stops.

PE= mgh
KE= 1/2 (m)(v^2)

You need to find how far the y vector of the swinging ball went which takes some trig. Make a line connecting the rope at 0 degrees and at 3 degrees. This makes a triangle of 30,75,75 angles. Bisect the 30 degree to get a right triangle of 15 75 90. Use sin(15)=opp/1 to get the length of .259 then multiply it by 2 to get the full length.

Make another triangle where you make a line of the x vector going right from the ball at 0 degrees. And then draw a line from the ball at 30 degrees down to make a 90 degree angle with the x vector. You should have another triangle of 15 75 90.
Sin(15)=opp/.518 solve for opp and you should get .134 that is the height that the ball rose.

Then just cancel out weight when equalling the two energies. 1/2(v^2)=gh

Sqrt(2gh) = v =0.512m/s

>3
30 degrees

h=r(1-cos(30))=0.134 m
v=sqrt(2gh)=1.62 m/s

It clearly shows two uplifted and eroded diapirs each with a dyke radiating from each one, which intersect further west than the map shows.
The north arrow is drawn a bit funny and they didn't quite finish writing the N
The diapirs also shouldn't have such big strike/dip measurements and they shouldn't have arrows on unless they're actually indicating lineation
Needs more outcrops, contours and could do with a key for the units
2/10 did you even go to your mapping classes

1.6209867024093

It's not a collision you dumbass it's a pendulum.