Aircraft drops package find vertical distance

[qwerty3]

An aircraft flying in a straight line at constant height h and speed U drops a package
on a target at ground level. The package is subject to air resistance whose magnitude
is kmv where v is the speed of the falling package, m its mass, and k > 0 a constant.
(a) Show that the vertical distance that the package falls in time t is

V(V/g(e^(-gt/V)-1)+t)

where V is the terminal velocity of the package, and find the horizontal distance
travelled in the same time t.
(b) Find the cartesian equation for the path of the falling package.

Hi i want to know what my constant k is??
when i use k=-g/V i get the wrong answer, but when i use k=g/V i get the negative of what they want:S:S

please help

 

[dynamicsolo]

I think you should not start off by trying to impose an expression for k. (You can see what the dimensions should be, but we don't know what might crop up dimensionlessly...)

Perhaps you should start with the two force equations, since this is a ballistics problem with air drag:

$$ma_x = -kmv_x$$
and
$$ma_y = mg - kmv_y$$

subject to the stated initial conditions
[ v_x(0) = U , v_y(0) = 0 , y(0) = 0, and call x(0) = 0 ]
and calling "downward" positive.

 

[qwerty3]

Well i took upwards as positive and so my mg is -mg.

 

[dynamicsolo]

Since you haven't shown the work you did, it is somewhat hard to judge why you aren't getting your signs to come out right. I suspect that you have not handled the differential equations and the initial conditions consistently in terms of signs.

 

[qwerty3]

ok i will show it its just that its hard to type it all up

Total force = -mgj -kmv
Newtons 2nd law a = -gj -kv => a + kv = -gj
Try integrating factor e^(int kdt)= e^(kt)


d/dt (e^kt v) = -ge^kt j
e^kt v = -g/k e^kt j + C

t=0 v=u => C = u + g/k j

v = (u + g/k j)e^-kt - g/k j

v=dr/dt

r = -1/k (u + g/k j) e^-kt - g/k tj + C'
r=0 t=0 => C' = 1/k (u + g/k j)

r = -1/k (u + g/k j) e^-kt - g/k tj + 1/k (u + g/k j)
r = (g/k^2 j + u/k)(1 - e^-kt) - g/k tj

u=ui as only horizotal speed

x(t) = u/k (1 - e^-kt)
y(t) = g/k^2 (1-e^-kt) -g/k t

and then when i work out what k is i don't get the answer as i get k=-g/V

 

[dynamicsolo]

ok i will show it its just that its hard to type it all up

Total force = -mgj -kmv

Ah, but there's a problem right here. If upward is positive and the object is falling, then with the weight force being downward and the drag force upward, this needs to be

$$ma_y = -mg + kmv_y$$

or just $$\frac{dv_y}{dt} = -g + kv_y$$

(Oh, and if you're going to write in unit vectors -- which isn't really necessary -- be sure to put them in across the board.)

The choice of axis makes things awkward, in that you must interpret v_y as negative throughout your solution. If you write the "correct" force equation, k must also be negative. If you use the equation you've written, k will be positive, but the velocity must still be treated as negative.

It really makes for less of a headache to call downward positive and work with

$$\frac{dv_y}{dt} = g - kv_y$$

but it should work either way if you treat all the signs of quantities consistently.

BTW, an integrating factor will work, but you could just solve this as a separable differential equation.

 

[qwerty3]

Actually what i want to know is that my way with vectors how do i split up the x and y component as with y the kmv is +ve

 

[LowlyPion]

i don't really get why the drag force is upwards please explain

Drag force retards speed. Gravity is accelerating speed. Whichever convention for direction you use, your equation should reflect that they have opposite signs.

 

[LowlyPion]

Actually what i want to know is that my way with vectors how do i split up the x and y component as with y the kmv is +ve

Not sure what you mean by "ve", but your x component of velocity comes from the initial speed of the plane. But of course that velocity slows as the package encounters the drag. Note that drag is proportional to velocity so as it slows the drag becomes less until it approaches zero as velocity approaches zero - in the x direction only.

The trajectory then is the combination of the x/y position equations as a function of time.

 

[qwerty3]

its ok thanks i get it now and +ve means positive :P sorry for the confusion