Air friction and acceleration in skydiver motion

[arkssd]

hello,

what would be the best way to approach this problem

A skydiver is jumping out of a plane at :

Initial velocity = 100 mph

Initial position = 2000 m

acceleration due to gravity = -9.81

I need help finding air friction at different positions, change in acceleration and change in velocity ?

i think u calculate air friction
A= f * v * v(abs) * cd

A= acceleration

f= atmospheric pressure

v= velocity

s= surface area

cd= drag coefficient

 

[Pyrrhus]

Start off with Newton's 2nd Law (For constant mass) in scalar form:

$$\sum_{i=1}^{n} F_{i} = ma$$

rewrite it as

$$\sum_{i=1}^{n} F_{i} = m \frac{dv}{dt}$$

or

$$\sum_{i=1}^{n} F_{i} = m \frac{d^{2} x}{dt^{2}}$$

And solve the ODEs formed

 

[quetzalcoatl9]

hello,

what would be the best way to approach this problem

A skydiver is jumping out of a plane at :

Initial velocity = 100 mph

Initial position = 2000 m

acceleration due to gravity = -9.81

I need help finding air friction at different positions, change in acceleration and change in velocity ?

i think u calculate air friction
A= f * v * v(abs) * cd

A= acceleration

f= atmospheric pressure

v= velocity

s= surface area

cd= drag coefficient

the general form of that equation looks familiar, but isn't it actually:

$$F(v) = c \rho A v^2$$

where [itex]A[/tex] is the surface area? Acceleration would then be:<br /> <br /> $$a(v) = \frac{c \rho A v^2}{m} = \frac{c \rho A}{m} ({\frac{dx}{dt}})^2$$<br /> <br /> integrating with respect to time will give the velocity as a function of time, but this is a 2nd order DE so you need to solve the DE and plug in your initial value conditions.[/itex]

 

[dextercioby]

Not really II-nd order.Use Newton's second law in vector and the definition of linear momentum.

Daniel.

 

[whozum]

Normalize your units first off.

 

[dextercioby]

There's no atmospheric pressure in the drag force,just the air's density...

Daniel.

 

[arildno]

the general form of that equation looks familiar, but isn't it actually:

$$F(v) = c \rho A v^2$$

where [itex]A[/tex] is the surface area? Acceleration would then be:<br /> <br /> $$a(v) = \frac{c \rho A v^2}{m} = \frac{c \rho A}{m} ({\frac{dx}{dt}})^2$$<br /> <br /> integrating with respect to time will give the velocity as a function of time, but this is a 2nd order DE so you need to solve the DE and plug in your initial value conditions.[/itex]

$Nope, this is false; air resistance always works in the opposite direction of the velocity; that's why you need the absolute value sign here.$