Question

Question #01f3b

Answer 1

Explanation:

Without drag, the acceleration vector is a constant vector and so the dynamics is symmetric between upward and downward motions.

But in the presence of drag, acceleration differs between the upward motion and downward motion. This is because the drag force always acts anti-parallel to the velocity vector.

Net Force & Acceleration: `\qquad vec F_{"net"} = vec F_g + vec F_d; \qquad vec a = vec F_{"net"}/m`

Upward Motion: When the object moves upward, both gravity and drag act downward.
`vec F_g = -mg; \qquad vec F_d = -\gammav^2; \qquad vecF_{"net"} = -mg -\gammav^2`
`vec a_{up} = -g - \gamma/mv^2 = -g[1 + (\gamma/(mg))v^2]` ... (1)

Downward Motion: When the object moves downward, gravity still acts downward but the drag force now acts upward, spoiling the symmetry.
`vec F_g = -mg; \qquad vec F_d = +\gammav^2; \qquad vecF_{"net"} = -mg +\gammav^2`
`vec a_{dn} = -g + \gamma/mv^2 = -g[1 - (\gamma/(mg))v^2]` ... (2)

Since the drag force acts opposite to the gravitational force during the downward motion, there is a possibility that they both might equilibrate, giving rise to a constant velocity. It is easy to calculate this terminal velocity and calculations look simple if we rewrite the equations in terms of this parameter

Terminal Velocity:
`vec F_g + vec F_d = vec 0; \qquad \rightarrow -mg + \gammav_t^2 = 0; \qquad v_t = \sqrt{(mg)/\gamma}`

The equations (1) and (2) for the accelerations in the upward and downward directions look simple if we use the terminal velocity parameter.

`veca_{up} = -g[1+(v/v_t)^2]; \qquad veca_{dn} = -g[1-(v/v_t)^2]`

Integrating acceleration once we can find the velocity as a function of time and by integrating the expression for velocity we get the position as a function of time. The following are the expressions for the velocity and positions.

Upward Motion: The mass `m` is projected upward with a initial velocity of `v_0`. The expressions for velocity and position look simple if we introduce additional parameters which are defined below.
`v(t) = v_t\tan[\chi_0 - t/\tau]; \qquad y(t) = v_t\tau\ln{\frac{\cos(\chi_0-t/\tau)}{\cos\chi_0}}`

`\chi_0 = \tan^{-1}(v_0/v_t); \qquad \tau = v_t/g;`

Downward Motion: The mass starts from a height `H` and falls down. Its velocity will asymptotically reach the terminal velocity value.
#v(t) = -v_t\tanh(t/\tau); \qquad y(t) = H - v_t\tau\ln[\cosh(t/\tau )]#

Note: Calculation details are scanned and uploaded.