Free Fall Motion

Core definition:

• Special case of 1D motion with constant acceleration where acceleration is due to gravity: $a = -g$ (if $+y$ is upward).
• $g = 9.8 \text{m/s}^2$ is a positive scalar magnitude; the negative sign indicates direction (downward).
• Critical reminder: Coordinate system dictates sign conventions. Always define $+y$ direction explicitly before solving.

Coordinate system setup:

• Upward throw (from ground level):
  • Initial velocity $v_0 > 0$ (upward), $a = -g$.
  • Position: $y(t) = v_0 t - \frac{1}{2} g t^2$
  • Velocity: $v(t) = v_0 - g t$
  • Key insight: Motion is symmetric; time to reach max height = time to fall back to start.
  • Max height: $h_{\text{max}} = \frac{v_0^2}{2g}$
  • Time of flight: $t = \frac{2v_0}{g}$
• Free drop (from height $h$):
  • It is easier to set $+y$ as downward, and origin as starting point
  • Initial velocity $v_0 = 0$, $a = g$
  • Position: $y(t) = \frac{1}{2} g t^2$ (with $y=0$ at ground).
  • Time to fall: $t = \sqrt{\frac{2h}{g}}$
  • Impact speed to reach ground $y = h$: $v = \sqrt{2gh}$ (magnitude; direction downward)
  • Critical reminder: $h$ must be positive (height above ground)

Advanced topic: Distance traveled in sequential fixed time intervals

• Problem definition:
  • It is not unique to free fall, but commonly defined in a free-fall setup
  • Given an object free fall with or without initial downward velocity, given a fixed time interval $\Delta t$, study the distance traveled in each time interval $\Delta y$
• General derivation:
  • Coordinate system:
    • $+y$ downward
    • Origin at the point of release.
  • At any two time points $t_1$, $t_2$, the coordinates the object reaches:
    • $y_1 = v_0 t_1 + \frac{1}{2}gt_1^2$
    • $y_2 = v_0 t_2 + \frac{1}{2}gt_2^2$
  • So distance traveled for this time interval $t_2 - t_1$ is:
    • $y_2 -y_1 = v_0(t_2 - t_1) + \frac{1}{2}g(t_2^2 - t_1^2)$
    • $y_2 -y_1 = (t_2 - t_1)(v_0 + \frac{1}{2}g(t_2 + t_1))$
  • So, written differently:
    • $\Delta y_{21} = (v_0 + \frac{1}{2}g(t_2 + t_1))\Delta t$
• What this means:
  • For fixed time interval $\Delta t$ is a constant, then $t_n$ linearly progresses with $n$, therefore $y_n$ is also linearly dependent on $n$.
  • For clarity, suppose $t_n =n\Delta t$, where $n = 0, 1, 2, ...$ In the above formulation, $t_2 = n\Delta t$, and $t_1 = (n-1)\Delta t$. Then we have:
    • $\Delta y_{n} = (v_0 + \frac{1}{2}g(2n-1)\Delta t)\Delta t$
  • Since $\Delta t$ is a constant, $\Delta y_n$ grows linearly with $\Delta t$:
    • If $v_0=0$, then $\Delta y_1 : \Delta y_2 : \Delta y_3 = 1: 3:5$
  • This can be used to accurately measure gravitational acceleration $g$, which is proportional to the linear coefficient of the $\Delta y_{n}$ vs $n$
• What if $\Delta y$ is constant:
  • Problem definition: Drop a ball from the top of the building, we want to see the time it takes to go through each level of the building, assuming no initial velocity, and each level has the same height.
  • In this case, we have: $v_0 = 0$ for simplicity, then we have:
    • $y =\frac{1}{2}gt^2$
    • $\Delta y_{21} = \frac{1}{2}g(t_2^2 - t_1^2)$
  • This means, for fixed $\Delta y$:
    • $t_n^2$ is an arithmetic sequence
    • $t_1^2 : t_2^2 : t_3^2 = 1:2:3$
    • $t_1 : t_2 : t_3 = 1: \sqrt 2: \sqrt 3$