Interestingly enough, there is nothing in canon about how the movement multiplier effects jumping or running-jumps. This was one of the first questions one of my players asked me after he started using cyberskates.

Vo = Initial Velocity (in m/s)

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The axis velocity components:

Vx = Vo*cos(a)

Vy = Vo*sin(a)

In the x direction the acceleration is zero so we can find how long it takes to go a distance d:

t = d/Vx = d/(Vo*cos(a))

In the y direction, the projectile undergoes an acceleration -g and does a displacement of zero because it lands at the same height it was launched from. In the third equation below, substitute the value for t from above and then do some pretty serious trig to solve for d and simplify.

d = Vo * t + (1/2) * a * t^2

0 = (Vo * sin(a)) * t + (1/2) * (-g) * t^2

0 = (Vo * sin(a) / (d / Vo * cos(a))) - (1/2) * g * (d / Vo * cos(a))

0 = d * tan(a) - ((d^2) * g) / (2 * Vo^2 * cos^2(a))

((d^2) * g) / (2 * Vo^2 * cos^2(a)) = d * tan(a)

((d * g) / (2 * Vo^2 * cos^2(a)) = tan(a)

d = (2 * Vo^2 * cos^2(a) * tan(a)) / g

d = (2 * Vo^2 * cos^2(a) / (sin(a) / cos(a)) / g

d = (2 * Vo^2 * cos(a) * sin(a)) / g

Using Trig Idendity that: sin(a) * cos(a) = (1/2) * sin(2*a)

ANSWER: Distance, d = (Vo ^2 * sin(2*a)) / g

The time it takes for the projectile to get to its maximum height is half the time it takes to get to d because of the symmetry of projectile motion. Take the time we got at the top and replace with d and then do some more simplification:

t = (1/2) * d / (Vo * cos(a))

t = (Vo^2 * sin (2*a)) / 2 * g * (Vo^2 * cos(a))

t = (Vo/2 * g) * (sin(2*a) / cos(a))

Using Trig Idendity that: sin(2*a) / cos(a) = 2 * sin (a)

t = (Vo / 2 * g) * 2 * sin(a)

ANSWER: Time, t = (Vo * sin(a)) / g

Now we plug in this value for t in the equation of motion for the y direction:

d = Vo * t + (1/2) * a * t^2

h = t * (Vo * sin(a)) + (1/2) * (-g) * t^2

h = (Vo * sin(a) * Vo * sin(a)) / g - (1/2) * g * (Vo * sin(a) / g) ^ 2

h = (Vo ^2 * sin^2(a)) / g - (Vo^2 * sin^2(a)) / 2 * g

ANSWER: Height, h = (Vo^2 * sin^2(a)) / 2*g

Max_Distance, d = (Vo ^2 * sin(2*a)) / g

Travel_Time, t = (Vo * sin(a)) / g

Max_Height, h = (Vo^2 * sin^2(a)) / 2*g

a = angle

Q = Quickness

Vo = Initial Velocity (in m/s) [for SR: m/s = Q x (running mult)/3]

A typical running-jump: Q=6, Vo=(6*3/3) = 6m/s, a=45.

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d = (Vo ^2 * sin(2*a)) / g

d = (6^2) * sin(2*45) / 9.8

d = (36 * 1) / 9.8

d = 3.67m running-jump distance.

h = (Vo^2 * sin^2(a)) / 2*g

h = ((6 ^ 2) * sin^2(45)) / 2 * 9.8

h = (36 * .5) / 19.6

h = .918m max height of jump

t = (Vo * sin(a)) / g

t = (6 * sin(45)) / 9.8

t = (6 * .707) / 9.8

t = .433 seconds from jump start to finish

d = 3.67m running-jump distance.

h = .918m max height of jump

t = .433 seconds from jump start to finish

Same running-jump as a Dwarf: Q=6, Vo=(6*2/3) = 4m/s, a=45.

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d = (Vo ^2 * sin(2*a)) / g

d = (4^2) * sin(2*45) / 9.8

d = (16 * 1) / 9.8

d = 1.63m running-jump distance.

h = (Vo^2 * sin^2(a)) / 2*g

h = ((4 ^ 2) * sin^2(45)) / 2 * 9.8

h = (16 * .5) / 19.6

h = .408m max height of jump

t = (Vo * sin(a)) / g

t = (4 * sin(45)) / 9.8

t = (4 * .707) / 9.8

t = .289 seconds from jump start to finish

d = 1.63m running-jump distance.

h = .408m max height of jump

t = .289 seconds from jump start to finish

CyberSkates in action: Q=6(10), Vo=(Q*6)/3= 20m/s, a=30.

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d = (Vo ^2 * sin(2*a)) / g

d = (20^2) * sin(2*30) / 9.8

d = (400 * .866) / 9.8

d = 35.3m running-jump distance.

h = (Vo^2 * sin^2(a)) / 2*g

h = ((20 ^ 2) * sin^2(30)) / 2 * 9.8

h = (400 * .25) / 19.6

h = 5.1m max height of jump

t = (Vo * sin(a)) / g

t = (20 * sin(30)) / 9.8

t = (20 * .5) / 9.8

t = 1.02 seconds from jump start to finish

d = 35.3m running-jump distance.

h = 5.1m max height of jump

t = 1.02 seconds from jump start to finish

It's harder to represent a standing jump using Quickness, I'd be more inclinded to use Strength, but either way: S=6, Vo=(S/3)=2m/s, a=90.

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d = (Vo ^2 * sin(2*a)) / g

d = (2^2) * sin(2*90) / 9.8

d = (4 * 0) / 9.8

d = 0m running-jump distance.

h = (Vo^2 * sin^2(a)) / 2*g

h = ((2 ^ 2) * sin^2(90)) / 2 * 9.8

h = (4 * 1) / 19.6

h = .204m max height of jump

t = (Vo * sin(a)) / g

t = (2 * sin(90)) / 9.8

t = (2 * 1) / 9.8

t = .204 seconds from jump start to finish

d = 0m running-jump distance.

h = .204m max height of jump

t = .204 seconds from jump start to finish

Problem with using a real-life approach would be making the jump wouldn't really be a test of skill as much as landing the jump. Since Vo for game purposes will equal Vt when landing I would think the TN should be based on the landing speed and conditions of the ground more that how high you jump.