Does increasing mass increase the terminal velocity of a parachute? Will a parachute that holds 10kg have a greater terminal velocity than a parachute that holds 5kg? I've heard conflicting information so i'm not sure if it does. If it does then why and how does it increase the terminal velocity?
 A: The terminal velocity occurs when the gravitational force pulling the object down is balanced by the drag force. The gravitational force ($Mg$), is proportional to the mass. The drag force depends mainly on the shape of the object, not on its mass, since it is determined by the flows of the surrounding fluid. The velocity dependence can be complicated, but is often something like $v^2$ for turbulent flows. Therefore, the terminal velocity will depend on the object's mass, as $v$ must be higher to balance $Mg$ for a larger $M$.
A: Adding some equations to Raghu's great answer:
The drag force on a body moving at a velocity $v$ in a fluid of density $\rho$ is given by
$$
F_d = \frac{1}{2} \rho v^2 S C_D \tag 1
$$
where $C_D$ is the coefficient of drag and $S$ is the corresponding area of the body based on the way $C_D$ is measured. There is some variation of $C_D$ for varying velocities, but as long as the flow stays well within the turbulent regime ($Re = \frac{\rho v d}{\mu} \gtrapprox 10^6$), this change isn't very large.

None of these terms depend on mass, assuming you didn't need to increase the area of your body in order to increase its mass. If we assume that the majority of drag comes from the parachute ($F_{d,{\rm parachute}} >> F_{d, {\rm payload}}$), which is usually the case), a change in the payload's area isn't going to cause a significant difference to our system anyway.
When terminal velocity is achieved, this drag force is cancelled out by gravity, so we have
$$
F_g = m g = F_d \tag 2
$$
\begin{align}
m g &= \frac{1}{2} \rho v^2 S C_D  \\
v &= \sqrt{\frac{2 m g}{\rho S C_D}} \\
v &\sim \sqrt{m}
\end{align}
We have the terminal velocity $v$ is proportional to the square root of the mass $m$.
A: Depends on the specifics of your question, if we have identical parachutes holding a mass of 5kg and 10kg, the 5kg bearing parachute will have a lower terminal velocity assuming the 10kg object has equal or greater density than the 5kg object. In this case, the force of air resistance would be about equal (assuming the 10kg object's density is about the same as the 5kg object) but the force pulling down on 10kg parachute would be greater resulting in a greater negative (towards the earth) force. Once a certain speed is reached, lets say 15kph, both parachutes will have equal force resisting their downward acceleration but the 10kg parachute has a greater force counteracting the drag force allowing it to accelerate to a greater top speed than the 5kg object.
Simple math explanation: assume the parachutes are weightless, the only mass in the systems are from the weights attached to the parachutes.
The 5kg bearing parachute will have a negative force of 5kg * 10kg/N = 50Newtons
the 10kg bearing parachute will have a negative force of 10kg * 10kg/N = 100Newtons
The "drag force" will be equal for both because the parachute is the same and the weight will be negligible compared to the parachute and will increase at a rate equal to the square of the velocity.
Lets say at 15kph the force is 50Newtons. This means that the 5kg parachute will have equal force acting up and down resulting in no net force while the 10kg parachute will still have 100N - 50N = 50N force acting down resulting in further negative acceleration until it finally reaches the speed where the drag force = 100N (in this example that would be ~21.21kph
If you were talking about a entirely different situation with a parachute rated for 5kg vs a different parachute rated for 10kg then it once again depends on the specifics and you would have to clarify further on the exact scenario you want to learn more about.
