i will try to make sense.

notes:

110" = 2.794meters

1/4" = .00635 meters

g=9.8 is the acceleration of gravity

m=mass, v=velocity, h=height from base

Fd=work done (force x distance)

(i.e. 4Newtons at top of the pole x .00635 meters to move plate)

Energy is always conserved. E = E

0.5mv^2 + mgh + Fd = 0.5mv^2 + mgh + Fd

(at bottom of the pole) = (at top of the pole)

0.5mv^2 + 0 (h=0) + 0 (no work done)=

0 (v=0) + m(9.8)(2.794meters) + (4Newtons)(.00635meters)

.5mv^2 = 27.3812m + .0254

mv^2 = 54.7624m + .0508

v^2 = 54.7624 + (.0508/m)

v = sqrt(54.7624 + (.0508/m))

this looks complicated: however, m (mass) is in kilograms.

Unless your bot is less than .1kg = .22lbs, that number will be negligible.

v approximates sqrt(54.7624) =7.4… m/s, i’m going to round up to 7.5 to be safe.

you need MINIMUM 7.5 m/s = 25 fps upward launching velocity if you were to launch an object from the base that does not touch the pole but effectively trips the sensor. (like a cylinder ring) This accounts for gravity not for air resistance (low) and friction (rubbing against the pole), if you had friction you should focus on programming a robot that actually climbs rather than launches.

concerning a push from a moving/climbing/(or launched) robot,

someone may have already answered this.

the energy transferred to the sensor is = Kinetic Energy (due to motion)

E = 1/2 m v^2 = Fd = (4N)(.00635meters) =.0254 Joules

So assuming your robot weighs about 2 pounds (1 kg) then

1/2 (1) v^2 = .0254

v^2 = .0508

v= .2253 m/s = 0.75 fps

The more the robot weighs, the less the required velocity.

In conclusion:

If you can get a robot moving and climbing the tower high enough in 10 seconds, you probably WILL trip the sensor.

If you launch something (unclear, GDC will clear it up soon) then you need about 25 fps from the base for that object to trip the sensor.

Good luck.