Question

Mass m is thrown horizontally in velocity v from the top of a tower (at height h). while the mass is moving, the air resistance force is equal to f=-bv (b is a constant). find the function of the position of the mass with respect to time

Answer 1

It looks like we can combine Kinematics and Newton's Second Law here.
(1) \[\Delta(x)=v(i)t+1/2at^2\]
(2)\[F=ma=-bv\]
By rearranging terms we can conclude

(3) \[a=-bv/m\]
Thus,
(4)\[x(t)=v(i)t+1/2(-bv/m)t^2\]

Answer 2

@Menono: Your solution is incorrect. The kinematic equations you used apply to constant acceleration only. Since the acceleration depends on the velocity, these equations are not valid here. Also, you failed to take in to account gravity. So, the velocity (and acceleration) are changing not only in magnitude, but direction as well.

Answer 3

I'll pick off from equation (3)
but will present it in a different way:
\[\frac{dv}{dt}=\frac{-b}{m}v\]
multiply by dt
\[dv=\frac{-b}{m}vdt\]
divide by v:
\[\frac{1}{v}dv=\frac{-b}{m}dt\]
do an integral on both sides
\[\int\limits_{v_i}^{v_f}\frac{1}{v}dv=\int\limits_{0}^{t}\frac{-b}{m}dt\]
we get:
\[ln\frac{v_f}{v_i}=-\frac{b}{m}t\]
or
\[\frac{v_f}{v_i}=e^{-\frac{b}{m}t}\]
\[v_x(t)=v_f=v_ie^{-\frac{b}{m}t}\]
after one more integral:
\[x(t)=\frac{mv_i}{b}(1-e^{-\frac{b}{m}t})\]

now the question is how much time does it take to the mass to drop from height h
tell me if you can't sort it out

Answer 4

Be aware that Marx has given you the position of the object wrt time in the x-coordinate only. To fully solve this one you need to work out y(t) in an analogous way as above (taking in to account the gravitational force as well as air resistance). To get you started, we have:\[\frac{dv _{y}}{dt}=-g-\frac{b}{m}v _{y}\]