The parachutist has a total mass of 90 kg and is in free fall at 6 m / s when she opens her 3

step 1 And this problem, we're considering a parachutist. I've given the information given in the probl

The parachutist has a total mass of 90 kg and is in free...

The parachutist has a total mass of $90 \mathrm{~kg}$ and is in free fall at $6 \mathrm{~m} / \mathrm{s}$ when she opens her 3 -m-diameter parachute. Determine the time for her speed to be increased to $10 \mathrm{~m} / \mathrm{s}$. Also, what is her terminal velocity? For the calculation, assume the parachute to be similar to a hollow hemisphere. The air has a density of $\rho_{a}=1.25 \mathrm{~kg} / \mathrm{m}^{3}$.

Key Concepts

Effective Area and Shape Considerations

The effective area in drag calculations is the projected area of the object that faces the oncoming flow. For objects with specific geometrical shapes, such as a hemisphere, determining this area accurately is crucial. The geometry not only influences the drag coefficient but also the effective area, both of which are essential for calculating the drag force and the consequent dynamics of the motion.

Differential Equation of Motion with Drag

When considering motion under gravity with drag, Newton's second law leads to a differential equation of the form m(dv/dt) = mg - F_drag. Because F_drag is a nonlinear function of velocity (often proportional to v²), solving for velocity as a function of time typically involves separating variables and integrating, leading to expressions that describe how speed approaches the terminal velocity over time.

Terminal Velocity

Terminal velocity is the constant speed reached by an object when the drag force equals the gravitational force, resulting in zero net acceleration. At this velocity, the object no longer accelerates because the forces acting on it (typically gravity and air resistance) are balanced. It is found by equating the gravitational force to the drag force and solving for velocity.

Drag Force

Drag force is the resistive force exerted by a fluid (such as air) on an object moving through it. It generally depends on the fluid's density, the object's speed, a drag coefficient that characterizes shape and surface roughness, and the effective cross?sectional area. The typical formula used is F_drag = ½ * ? * v² * C_d * A, which describes how the force increases with the square of the velocity and depends directly on the medium and object properties.

Transcript

00:01 And this problem, we're considering a parachutist.

00:03 I've given the information given in the problem here.

00:05 We have a mass of 90 kilograms.

00:07 Our initial speed is 6 meters per second.

00:09 Our final speed is 10 meters per second.

00:12 Our diameter of our parachute is 3 meters.

00:15 We can assume the parachute to be a hollow hemisphere.

00:19 Our viscosity is given right here.

00:21 I'll look that up, and we're given the density of 1 .25 kilograms per meter cubed.

00:26 We're asked first to determine the time for the speed to be increased from 6 to 10 meters per second and to do second to determine the terminal velocity.

00:37 Okay, let's begin.

00:40 First, let's do our sum of our y forces and that will be negative m over.

01:10 Okay, and the characteristic length, so let's call this equation one.

01:16 And then our characteristic length for hollow hemisphere is diameter d, a reynolds number.

01:32 It will be this and this will equal 1 .25.

01:50 Times let's do 10 meters per second and three meters for a diameter divided by my 17 .55 times 10 to the negative 6th and this will equal 2 .136 times 10 to the minus 6 which is greater than 10 to the 4th so we can use since a reynolds number is greater than 10 to the 4th we can use a table to get the drag coefficient.

02:46 Okay.

02:47 So our drag coefficient will be 1 .4.

02:53 Now, our drag is defined with speed, density, and projected area.

03:03 Our drag is our area of p, a parachute will be d squared...

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