# 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?

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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.

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You did it right, but in order to be sure, you should do the calculation multiple times, each time doing more in your head, until you internalize everything (including the calculations--- set g=10 to do the arithmetic) and then you can introspect to see that it is right. The level of detail you show is more than is necessary. Once you repeat it enough, the solution becomes this: an object with velocity v will go up to maximum in v/g time, and travel (v/2) * (v/g) = v^2/2g distance up. You are effectively reducing g by 40% without changing v, so you are increasing the height by 40%. – Ron Maimon Jul 5 '12 at 2:38