Motion Under Gravity

Near the Earth's surface, every freely falling body experiences a constant acceleration $g \approx 9.8\,\text{m/s}^2$ downward (we often use $10\,\text{m/s}^2$ for simplicity). All previous kinematic equations apply with $a = -g$ (taking up as positive).

Concept

Sign Convention

Take upward as positive. Then:

$g$ acts downward, so $a = -g$ always (regardless of motion direction).
An object thrown up has initial velocity $u > 0$ ; an object dropped has $u = 0$ .

Equations

With $a = -g$ , the three kinematic equations become: $v = u - gt,\qquad y = ut - \tfrac{1}{2}gt^2,\qquad v^2 = u^2 - 2gy.$ Here $y$ is the upward displacement from the release point.

Key Results

Object thrown upward with speed $u$ :

Time to maximum height: $t_{\text{up}} = u/g$ .
Maximum height: $H = u^2/(2g)$ .
Time of flight (back to start): $T = 2u/g$ .
Velocity at return: $-u$ (same speed, opposite direction).

Object dropped from height $h$ :

Time to hit ground: $t = \sqrt{2h/g}$ .
Final velocity: $v = \sqrt{2gh}$ .

Symmetry of Free Motion

For an object launched straight up:

Time going up = time coming down (to the same height).
Speed at any given height is the same on the way up and on the way down (only direction differs).
This is exactly true only when air resistance is neglected.

Worked Example

Q: A ball is thrown upward from a 20 m cliff with initial speed $15$ m/s. Take $g = 10$ m/s², up as positive. Find (a) maximum height above the cliff, (b) total time before hitting ground, (c) speed at ground.

Solution:

(a) At max height, $v = 0$ : $0 = 15^2 - 2(10)H \Rightarrow H = 11.25$ m above cliff.

(b) Position relative to cliff top: $y = 15t - 5t^2$ . Ground is at $y = -20$ : $-20 = 15t - 5t^2 \Rightarrow 5t^2 - 15t - 20 = 0 \Rightarrow t^2 - 3t - 4 = 0.$ $t = (3 + \sqrt{9+16})/2 = (3+5)/2 = 4$ s.

Sanity check via $v^2 = u^2 - 2gy$ : $v^2 = 225 - 2(10)(-20) = 225 + 400 = 625$ , $|v| = 25$ m/s ✓.

Common Confusions

" $g$ is positive going down." — Sign depends on your convention. With up positive, $a = -g$ throughout.
"Acceleration is zero at the top." — Velocity is zero; acceleration is still $-g$ .
"Heavier objects fall faster." — Neglecting air resistance, all objects accelerate the same.
Using $u = 0$ for thrown objects — only set $u = 0$ for objects dropped from rest.

Key Takeaways

$g \approx 9.8$ m/s² (often 10 m/s² in problems) downward.
Same kinematic equations with $a = -g$ (up positive convention).
Max height $= u^2/(2g)$ , time of flight $= 2u/g$ for vertical projection.
Free fall: $v = \sqrt{2gh}$ .
Acceleration never becomes zero during flight, even at peak.