Air resistance - find Vx(t) and Vy(t)

[Carnivroar]

Suppose at time t= 0 the football is kicked with initial velocity components Vxo and Vyo. Solve the equations above to find Vx(t) and Vy(t). You can make use of the formulas we worked out in class.

"Equations above":
ax = -kvx
ay = -g-kvy

formula worked in class:

v(t) = (g/k)(1-e^-k(t-to)) + Voe^-k(t-to)

so...

Vx(t) = Vocosθe^-kt

Vy(t) = (9.8m/s^2/k)(1-e^-kt) + Vysinθe^-kt

Is this correct?

 

[Carnivroar]

Next problem is to integrate that to get a formula for x(t) and y(t)

Assuming the above is correct, here's what I got:

x(t) = -V₀sin\thetae^-k(t-t_o)/k +V₀sin\theta/k

y(t) = -ge^-k(t-t₀)/k² + V₀cos\thetae^-k(t-t₀) + g/k² - V₀cos\theta/k

 

[Carnivroar]

Please help [URL]http://img.photobucket.com/albums/v253/Hatsuharu1399/gdkifmqq.gif[/URL]

 

[Carnivroar]

Come on guys

Let me rephrase my question

How to I include air resistance (k) into a projectile motion problem?

In this problem, k = 0.1/s

so do I stick that into the acceleration equation to get a new acceleration?

a_x = -kV_x
a_x = -0.1/s * V_x

a_y = g - kvy
a_y = -9.8m/s^1 - 0.1 * V_y

This part seems okay, but what do I use for Vx and Vy? Is it the initial velocities?

In this problem I was given V_x0 = 12 m/s and a_x0 = 10 m/s

So if I use that, I get these constant velocities

a_x = -1.2 m/s^2
a_y = -10.8 m/s^2

 

[SammyS]

Please help [PLAIN]http://img.photobucket.com/albums/v253/Hatsuharu1399/gdkifmqq.gif[/QUOTE]
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