How do I factor in multiple forces into these Newtonian mechanics equations? 
A person jumps.
The person weighs $25$kg (irrelevant?).
Just after jumping, their velocity is 5m/s (positive is taken as up).
Gravity is taken as $-9.80665\text{ m/sec}^2$.
Using $(v_1-v_0)/g$, I know that it will take $0.509858106488964$ seconds
  for the person to have no velocity (before accelerating downwards).
  This is $t$.
Using ($v_0t+.5gt)$, I know their maximum height should be
  $0.049290532444821m$.

Now, say I attach a helium balloon to that person.
It provides $10$kg of lift.
How do I translate that into a force that can then be factored in to those two equations?
 A: So, i believe i've figured it out:
To get the acceleration of gravity and the lift of the balloon working together, we convert them both to forces (in newtons) and then just add/subtract and get the resultant force.
F = A  
F = 25*-9.81  
Gravity = -245.25 N

F = MA  
F = 10*9.81 (9.81 because to counteract gravity it must be equal but opposite)  
Lift = 98.1 N

Resultant Force = 98.1 - 245.25
Resultant Force = -147.15 Newtons (down, as indicated by the negative)

Then we can reverse that to get the resultant acceleration:
F = MA
-147.15 = 25A
    /25
A = -5.886 (m/s^2)

Then we use the SUVAT equations to get the time and displacement to peak.  
Remember:
S = Target (we're trying to figure out)
U = 5 m/s (stating Velocity)
V = 0 (no movement at peak)
A = -5.886 m/s^2 (we just figured this one out)
T = Target (another one)
V = AT + U
0 = 5.668*T + 5
5 = 5.668*T
T = 0.882145378 (seconds)

They do stay up longer (now i can sleep).
After i figure out if they go higher too, or just decelerate more slowly.
S = ((V + U)/2)*T  
S = ((0 + 5)/2)*T  
S = 2.5*T  
S = 2.5*0.882145378
S = 2.20536345 (meters)  

This is much higher than the original (without the balloon), which worries me i've done it wrong, but i can't see where.
