I was under the impression we do indeed limit LS as well with the same limitations as KE, as far as im aware we've enforced this as such so not like SoL or FTL Lifting strength calcs same as KE. Regardless though we do have more than 1 way of doing jumping calculations that can be done without the regular speed portion
For SoL and above velocities, sure, but timeframes can get arbitrarily short, rendering acceleration arbitrarily high.
Lets just say we jumped 1000m into the air, We will assume 45% body height for the leg work of the force in the jump (this is just assuming legs are 45% of the total height)
62kg * 1000m * 9.81m/s = 608220 joules/ (1.71m*45%) = 790409.357N/9.81 = 8216.0370667kg (Class 10)
I'd imagine this is a fine work around for the concern in question
That's a bit of an underestimate for jumping, actually.
Ignoring air friction, the time needed for gravity to move you a certain vertical height is sqrt((2*h)/g), since the height is 1000 meters, and it's in Earth's gravity, that gives a time of ~14.28 seconds. Since jumping is functionally falling from a height in reverse, you would need a speed that reaches 0 velocity after 14.28 seconds to jump 1000 meters high, which means a velocity of ~140 m/s.
One would need to reach that velocity by the time their feet leave the ground. Assuming a leg length of 0.8 m, and halving that due to human fallibility with jumping technique and due to shins not being bendable, means one needs to accelerate to that speed before moving 0.4 m
Assuming linear acceleration, their average speed over that time will be half of their final speed, so 70 m/s, which means that they accelerated to that velocity in 1/175th of a second. Giving an acceleration of 24,500 m/s^2.
Multiplied by the weight of 62 kg, that gives a force exerted of 1,519,000 Newtons, or ~155,000 kgf,
Class K.
Yours ended up so much lower since you divided by 9.81^2 in the last step, which is not the correct conversion between Newtons and kilogram force (unless one of those 9.81 divisions was part of the method, in which case that method is just a substantial underestimate).
EDIT: We could even collapse this all into one thing to hide the use of velocity; F = (((sqrt(2*h/g))*g)^2)*m/(2*L), where F is force, h is final height, g is the gravitational constant, m is the mass, and L is the length the force was done over (typically, the difference in height between initiating the jump, and losing contact with the ground).