Free fall — g = 9.8 m - s², sign conventions: Exercises

Before we start, one reminder of what every symbol means, so nothing sneaks in undefined:

Figure — Free fall — g = 9.8 m - s², sign conventions

The figure above is the mental picture behind every problem: an object on a vertical line, one arrow showing the positive direction you chose, and a fixed downward arrow for gravity. Everything else is just plugging numbers into the three boxed equations.

Level 1 — Recognition

L1.1 — Which sign is $a$ ?

A rock is dropped from a bridge. You decide to call downward the positive direction. What is the acceleration $a$ (with its sign)?

Recall Solution

Setup: Down is positive. Gravity points down, which is now the positive way. Therefore $a = + g = + 9.8 m/s^{2}$ . Why: acceleration always physically points toward Earth; the sign only records whether that agrees or disagrees with the axis you picked. Down positive ⟹ they agree ⟹ plus.

L1.2 — Reading off the velocity equation

Using up positive, a ball is thrown up at $v_{0} = + 12 m/s$ . Write the equation for its velocity $v$ at time $t$ , and state the velocity at $t = 0$ .

Recall Solution

Setup: Up positive ⟹ $a = - g = - 9.8$ . Plug into $v = v_{0} + a t$ : $v = 12 - 9.8 t .$ At $t = 0$ : $v = 12 - 0 = + 12 m/s$ — exactly the launch speed, pointing up. Good sanity check.

L1.3 — What is true at the very top?

A ball thrown straight up reaches its highest point. Fill in: at that instant, the velocity is ____ and the acceleration is ____.

Recall Solution

Velocity $v = 0$ (the ball momentarily stops moving vertically). Acceleration $a = - g = - 9.8 m/s^{2}$ — gravity never switches off. Why $v = 0$ but $a \neq = 0$ : velocity being zero means "not moving right now"; acceleration being nonzero means "still changing velocity", which is why the ball starts falling back.

Level 2 — Application

L2.1 — Dropped from a height

A stone is dropped from rest from a height of $45 m$ . How long does it take to hit the ground?

Recall Solution

Setup: Up positive, $y_{0} = 45$ , ground at $y = 0$ , $v_{0} = 0$ ("dropped"), $a = - 9.8$ . Use $y = y_{0} + v_{0} t + \frac{1}{2} a t^{2}$ : $0 = 45 + 0 - \frac{1}{2} (9.8) t^{2} ⟹ 4.9 t^{2} = 45 ⟹ t^{2} = 9.184.$ $t = 9.184 = 3.03 s .$ We keep the positive root because time runs forward.

L2.2 — Impact speed

For the same $45 m$ drop, find the speed when it hits the ground.

Recall Solution

Fastest route: $v^{2} = v_{0}^{2} + 2 a Δ y$ . Here $Δ y = y - y_{0} = 0 - 45 = - 45$ (it moved down, opposite to our positive axis), $v_{0} = 0$ , $a = - 9.8$ : $v^{2} = 0 + 2 (- 9.8) (- 45) = 882 ⟹ v = - 882 = - 29.7 m/s .$ Why negative root: the stone is moving downward at impact, opposite our positive (up) direction. Its speed (magnitude) is $29.7 m/s$ .

L2.3 — Thrown down

A ball is thrown downward at $8 m/s$ from a $60 m$ cliff. Find its impact speed.

Recall Solution

Setup — down positive (everything falls one way, so all numbers stay positive → fewer slips): $a = + 9.8$ , $v_{0} = + 8$ , $Δ y = + 60$ . $v^{2} = v_{0}^{2} + 2 a Δ y = 8^{2} + 2 (9.8) (60) = 64 + 1176 = 1240.$ $v = 1240 = 35.2 m/s (downward) .$

Level 3 — Analysis

L3.1 — Max height and total flight time

A ball is thrown straight up at $v_{0} = 20 m/s$ from ground level. (a) Find the maximum height. (b) Find the total time until it returns to the ground.

Recall Solution

Setup: Up positive, $y_{0} = 0$ , $v_{0} = + 20$ , $a = - 9.8$ . (a) Max height. Let $H$ be the maximum height — that is, the displacement from the launch point up to the top, so $H = Δ y = y_{top} - y_{0}$ . Why we can find it without time: at the top $v = 0$ (the ball momentarily stops), and the time-free equation $v^{2} = v_{0}^{2} + 2 a Δ y$ links velocity to displacement directly. Set $v = 0$ and $Δ y = H$ : $0 = 2 0^{2} + 2 (- 9.8) H ⟹ H = \frac{400}{19.6} = 20.4 m .$ (b) Total time: the path is symmetric — time up equals time down. Time to top from $v = v_{0} + a t$ : $0 = 20 - 9.8 t_{up} ⟹ t_{up} = \frac{20}{9.8} = 2.04 s .$ $t_{total} = 2 t_{up} = 4.08 s .$

The plot below shows this whole flight. Look at it visually: the curve rises, flattens at the yellow dot (where the tangent is horizontal — that is $v = 0$ ), then falls back mirror-symmetrically. The dashed vertical line at $t_{up} = 2.04 s$ splits the parabola into two equal halves — that visual symmetry is exactly why time up equals time down, and why the peak sits at the dead centre of the flight.

L3.2 — Velocity at a given height

For that same throw ( $v_{0} = 20$ ), what is the velocity when the ball is at height $y = 15 m$ , on the way up? What about the same height on the way down?

Recall Solution

Use $v^{2} = v_{0}^{2} + 2 a Δ y$ with $Δ y = 15 - 0 = 15$ : $v^{2} = 400 + 2 (- 9.8) (15) = 400 - 294 = 106 ⟹ v = \pm 10.3 m/s .$ The equation gives two signs because the ball passes $15 m$ twice:

On the way up: $v = + 10.3 m/s$ (moving up).
On the way down: $v = - 10.3 m/s$ (same speed, moving down). This is the up–down symmetry, popping out of the algebra automatically. On the figure above, a horizontal line at $y = 15 m$ would cut the parabola in two places — one on the rising half, one on the falling half — which is the same two-crossing fact seen graphically.

Level 4 — Synthesis

L4.1 — Two balls, staggered release

Ball A is dropped from rest from a $50 m$ tower. One second later, ball B is dropped from the same point. When A has fallen for $3 s$ , how far apart are the two balls?

Recall Solution

Setup — down positive, origin at the release point. Distance fallen $= \frac{1}{2} g t^{2}$ where $t$ is each ball's own fall time.

Ball A has been falling for $t_{A} = 3 s$ : $y_{A} = \frac{1}{2} (9.8) (3)^{2} = 4.9 \times 9 = 44.1 m$ .
Ball B started $1 s$ later, so at this instant $t_{B} = 3 - 1 = 2 s$ : $y_{B} = \frac{1}{2} (9.8) (2)^{2} = 4.9 \times 4 = 19.6 m$ .
Separation $= y_{A} - y_{B} = 44.1 - 19.6 = 24.5 m$ . (Check A is still airborne: it hits ground when $\frac{1}{2} g t^{2} = 50 \Rightarrow t = 3.19 s > 3$ . Good.) Insight: the gap between two objects dropped a fixed time apart keeps growing, because the earlier ball is always moving faster — the head start in time becomes an ever-larger head start in distance.

L4.2 — Meeting in mid-air

From the ground, a ball is thrown up at $v_{0} = 25 m/s$ . At the same instant, a second ball is dropped from rest from a height of $40 m$ directly above. At what time do they meet, and at what height?

Recall Solution

Setup: Up positive, common origin at the ground, $t = 0$ for both.

Thrown ball: $y_{1} = 0 + 25 t - \frac{1}{2} (9.8) t^{2}$ .
Dropped ball: $y_{2} = 40 + 0 - \frac{1}{2} (9.8) t^{2}$ . They meet when $y_{1} = y_{2}$ . The $- \frac{1}{2} (9.8) t^{2}$ terms are identical, so they cancel: $25 t = 40 ⟹ t = \frac{40}{25} = 1.6 s .$ Why those gravity terms cancel — the physics, not just the algebra: both balls feel the exact same acceleration $- g$ (mass never entered — see Newton's Second Law). So gravity drags both downward by the identical amount $\frac{1}{2} g t^{2}$ at every instant. If you watch the gap between them, that shared falling motion contributes nothing — it shifts both balls equally. The gap therefore closes purely at the throw speed $25 m/s$ , exactly as if there were no gravity at all for the relative motion. Starting $40 m$ apart and closing at $25 m/s$ , they meet after $40/25 = 1.6 s$ — which is the equation above, now with a picture behind it. Height: $y = 40 - \frac{1}{2} (9.8) (1.6)^{2} = 40 - 4.9 \times 2.56 = 40 - 12.5 = 27.5 m$ .

Level 5 — Mastery

L5.1 — Ball thrown up from a moving platform… then dropped past the edge

A ball is thrown upward at $v_{0} = 15 m/s$ from the edge of a cliff that is $30 m$ above the ground below. Find: (a) the total time from launch until it lands on the ground, (b) its velocity just before landing.

Recall Solution

Setup: Up positive, origin at the launch point (cliff edge), so the ground is at $y = - 30$ . $v_{0} = + 15$ , $a = - 9.8$ . (a) We need the moment the ball reaches the ground, i.e. $y = - 30$ . Substitute into $y = y_{0} + v_{0} t + \frac{1}{2} a t^{2}$ with $y_{0} = 0$ : $- 30 = 15 t - \frac{1}{2} (9.8) t^{2} ⟹ - 30 = 15 t - 4.9 t^{2} .$ Why rearrange into $4.9 t^{2} - 15 t - 30 = 0$ : this is a quadratic in $t$ , and the quadratic formula only works once every term is on one side in the standard shape $A t^{2} + B t + C = 0$ . Move all terms to the left (which flips their signs) to get $A = 4.9$ , $B = - 15$ , $C = - 30$ . Solve with $t = \frac{- B \pm B ^{2} - 4 A C}{2 A}$ : $t = \frac{15 \pm 1 5 ^{2} + 4 ( 4.9 ) ( 30 )}{2 ( 4.9 )} = \frac{15 \pm 225 + 588}{9.8} = \frac{15 \pm 813}{9.8} = \frac{15 \pm 28.5}{9.8} .$ The two roots are $t = \frac{43.5}{9.8} = 4.44 s$ and $t = \frac{- 13.5}{9.8} = - 1.38 s$ . Why we keep the $+ 4.44 s$ root and discard $- 1.38 s$ : the equation describes the ball's height as a smooth parabola in time. It crosses the ground level $y = - 30$ at two instants — but one of them ( $t = - 1.38 s$ ) is before we released the ball, an imaginary "past" where the parabola, if run backward, would also have passed that height. Physically the motion only began at $t = 0$ , so only the future crossing is real. Hence $t = 4.44 s$ . (b) Velocity at landing from $v = v_{0} + a t$ : $v = 15 - 9.8 (4.44) = 15 - 43.5 = - 28.5 m/s .$ Negative ⟹ moving downward; impact speed $= 28.5 m/s$ . Cross-check with the time-free equation: $v^{2} = 1 5^{2} + 2 (- 9.8) (- 30) = 225 + 588 = 813$ , so $∣ v ∣ = 813 = 28.5 m/s$ ✓ — it agrees perfectly.

The picture below makes the "broken symmetry" concrete. Trace the blue curve: it rises to the yellow dot (the top, where the tangent is flat and $v = 0$ ), falls back through the launch height (the dashed white line), and then keeps going below it, all the way down to the green ground line $30 m$ lower. Notice how the falling stretch is visibly longer than the rising stretch — that extra length is the ball's journey below its starting point, and it is exactly why the simple "twice the time to the top" rule fails here.

L5.2 — Where the naive "up = down time" symmetry breaks

For the L5.1 cliff throw, a student claims "time going up = time coming down, so the flight is symmetric." Explain quantitatively why the total time is not simply twice the time to the top, and give both numbers.

Recall Solution

Time to the top (where $v = 0$ ): $t_{up} = v_{0} / g = 15/9.8 = 1.53 s$ . If the ball returned to the launch height, the trip down would also take $1.53 s$ , giving $3.06 s$ . But the ground is $30 m$ below the launch point — the ball keeps falling past its starting level. So the descent is longer: $t_{down} = t_{total} - t_{up} = 4.44 - 1.53 = 2.91 s .$ The clean up–down symmetry only holds between equal heights (look again at the figure: the mirror symmetry lives entirely above the dashed launch line). Here start and finish heights differ by $30 m$ , so the descent leg is bigger. The symmetry is a special case, not a law.

Connections

Equations of Motion (constant acceleration) — the three equations powering every solution here.
Vectors and sign conventions — line one of every problem: pick a positive direction.
Projectile Motion — these vertical calculations are the $y$ -component of any projectile.
Newton's Second Law — the reason $a = g$ for every mass.
Air Resistance and Terminal Velocity — remove the "ignore air" assumption and these numbers change.

Recall Self-test checklist

Did you, for every problem: (1) state up-vs-down positive first, (2) give $a$ the correct sign, (3) match the sign of $Δ y$ to that axis, (4) keep the physically correct root of any square root or quadratic? Answer ::: If yes to all four, sign errors — the #1 source of lost marks in free fall — are behind you.