Throw a ball straight up and it is in Free Fall the moment it leaves your hand: on the way up, at the top and on the way down. One acceleration, $9.8\ \text{m/s}^2$ downward, covers the whole flight, so one set of equations does too.
Key idea
With up positive, $a = -9.8\ \text{m/s}^2$ for the entire flight. At the top, $v = 0$ for an instant, but the acceleration is still $9.8\ \text{m/s}^2$ downward.
Why it works
Gravity doesn't switch off at the top. The velocity is changing there from up to down, and a changing velocity means an acceleration. If $a$ really were zero at the top, the ball would hover there forever. The flight is also symmetric: going up loses $9.8\ \text{m/s}$ each second and coming down gains it back, so the trip up and the trip down take equal times and the ball returns to your hand at its launch speed.
Worked examples
Worked example: How high and how long
A ball is thrown straight up at $19.6\ \text{m/s}$. Find the time to the top and the maximum height.
- At the top $v = 0$: $0 = 19.6 - 9.8t$, so $t = 2\ \text{s}$.
- $v^2 = v_0^2 + 2a\Delta y$: $0 = 19.6^2 - 2(9.8)\Delta y$, so $\Delta y = \dfrac{384.16}{19.6} = 19.6\ \text{m}$.
Answer: $2\ \text{s}$ to the top; $19.6\ \text{m}$ high
Worked example: Past the top
Same throw. What is the velocity $3\ \text{s}$ after release?
- $$v = 19.6 + (-9.8)(3) = -9.8\ \text{m/s}.$$
- Negative: the ball passed the top at $2\ \text{s}$ and is now falling.
Answer: $-9.8\ \text{m/s}$
Worked example: Landing below the start
From a cliff edge $25\ \text{m}$ high you throw a stone up at $10\ \text{m/s}$. When does it hit the ground below?
- Up positive, origin at your hand: the ground is at $\Delta y = -25\ \text{m}$.
- $-25 = 10t - 4.9t^2$, or $4.9t^2 - 10t - 25 = 0$.
- Quadratic formula: $t = \dfrac{10 \pm \sqrt{100 + 490}}{9.8}$. The positive root is $t \approx 3.50\ \text{s}$.
Answer: about $3.50\ \text{s}$
Common mistake
Setting $a = 0$ at the top because $v = 0$ there. Zero velocity and zero acceleration are different things: a ball at the top has no speed but is already gaining downward speed. Check on the graph: the $v$–$t$ line crosses zero with the same slope it always had.