1.5 Free fall and projectiles

Anything flying through the air with only gravity acting on it follows the same rules: steady sideways motion, plus free fall up and down. Split the motion into two easy problems.

Little picture stories

Press Next (or Play) to walk through each story one small step at a time. The dots left behind show where the ball was, at equal times.

1. Dropping a ball: faster and faster

2. Drop one, throw one sideways

From the same height at the same moment, one ball is dropped and one is thrown sideways. Which lands first?

3. Two motions at once: sideways steady, down faster and faster

A ball flies through the air (no air resistance). Its sideways speed:

4. Check yourself

Think of your answer first, then tap to see it.

a) Lily drops a ball and Sam throws one sideways, from the same height at the same moment. Which lands first?

Show answer

They land together. Both start with no up-or-down speed and gravity pulls both the same way. Sideways speed only changes where a ball lands, not when.

b) A dropped ball: how much does its speed grow every second?

Show answer

About 9.8 m/s every second (a = 9.8 m/s2 down). After 1 s about 9.8 m/s, after 2 s about 19.6 m/s.

c) A ball is thrown straight up. At the very top, what are its velocity and acceleration?

Show answer

Velocity 0 for an instant. Acceleration still 9.8 m/s2 down: gravity never switches off, so the velocity keeps changing, from up to down.

Already know this?

Three quick questions. Get all three right on the first try and you can skip ahead.

Up is positive. The acceleration of a ball in free fall (going up or down) is:

A ball is thrown horizontally off a cliff. Its horizontal velocity:

A stone is dropped from rest. How far does it fall in the first second?

Drop one, throw one

Read the story as text

Two friends stand on a balcony 20 m above the ground. At the same instant, one lets go of a tennis ball and the other throws an identical ball straight out sideways, fast.

The thrown ball flies far out over the lawn. The dropped ball goes straight down. Most people guess the dropped ball lands first because it has "less distance to go".

In fact both balls hit the ground at the same moment. Why does moving sideways not change how long a ball takes to fall, and how do we predict where a thrown ball lands?

The idea: two motions at once

In pictures

  • A dropped ball falls faster and faster: the dots it leaves get further apart. Its speed grows about 10 m/s every second.
  • Thrown sideways, a ball keeps the same sideways speed the whole way (equal sideways steps), while it falls exactly like a dropped ball.
  • So a dropped ball and a ball thrown sideways from the same height land at the same moment.
  • Thrown up, a ball slows, stops for an instant at the top, and comes back down. Gravity pulls down the whole time, even at the top.

Below is the same idea in words, and then with numbers.

Free fall

When only gravity acts (we ignore air resistance), every object near Earth's surface has the same acceleration: 9.8 m/s² downward. Heavy or light, going up, at the top, or going down: the acceleration is always the same.

We usually pick up as positive, so a = −9.8 m/s². Then the Unit 1 equations work as they are:

v = v0 + a t    y = y0 + v0 t + ½ a t²    v² = v0² + 2a(y − y0)

At the very top of a throw, the velocity is zero for an instant, but the acceleration is still −9.8 m/s². The velocity is still changing, from + to −.

Vector components

A velocity at an angle θ above horizontal can be split into a horizontal part and a vertical part:

v0x = v0 cos θ     v0y = v0 sin θ

Projectiles: x and y are independent

That is why the two balls in the story land together: both start with vy = 0 at the same height, so their vertical motions are identical. The sideways speed only changes where the thrown ball lands.

a = 9.8 m/s² down (even at the top) vy = 0, vx same same v_x arrow every time; v_y shrinks, flips, grows

Same idea with numbers (1): thrown sideways off a cliff

A ball is thrown horizontally at 12 m/s from the top of a 20 m cliff. Up is +y, outward is +x. Find the time in the air, how far out it lands, and its speed at impact.

Show all steps as text
  1. List what we know. x: vx = 12 m/s, ax = 0. y: v0y = 0, ay = −9.8 m/s², Δy = −20 m. "Horizontally" means all of the starting speed is in x; nothing in y.
  2. Find t from the y motion: −20 = 0·t + ½(−9.8)t², so t² = 40 ÷ 9.8 = 4.08 s², and t = 2.02 s. y is the direction that decides when the flight ends (when it reaches the ground).
  3. Use the same t in x: Δx = 12 m/s × 2.02 s = 24.2 m. x and y share the same clock.
  4. Vertical velocity at impact: vy = 0 + (−9.8)(2.02) = −19.8 m/s. Horizontal is still 12 m/s. Gravity only changes the y part.
  5. Speed at impact: √(12² + 19.8²) = √536 = 23.2 m/s, at tan⁻¹(19.8/12) = 58.8° below horizontal. Combine the components with Pythagoras to get the size of the velocity.

Same idea with numbers (2): launched at an angle

A ball leaves level ground at 20 m/s, 30° above horizontal. Find the time to the top, the maximum height, the total time and the range.

Show all steps as text
  1. Components: v0x = 20 cos 30° = 17.3 m/s, v0y = 20 sin 30° = 10.0 m/s. Split the launch velocity so each direction becomes a 1D problem.
  2. Time to the top: vy = 0 there, so 0 = 10.0 − 9.8 t, ttop = 1.02 s. At the top the ball stops rising for an instant; vy = 0 but vx is still 17.3 m/s.
  3. Max height: 0² = 10.0² + 2(−9.8)h, so h = 100 ÷ 19.6 = 5.10 m. This is the v² equation; no time needed.
  4. Total time on level ground: up and down take equal times, so t = 2.04 s. Check: y = 10.0(2.04) − 4.9(2.04)² = 20.4 − 20.4 = 0. ✓
  5. Range: Δx = 17.3 m/s × 2.04 s = 35.3 m. Constant horizontal velocity for the whole flight.

Play: free fall and projectiles

positionvelocityacceleration

Sim 1: from the top of a 45 m building

Up is positive. Set the throw velocity: positive throws up, negative throws down, zero just drops. Change the mass too, and watch what happens (nothing, without air resistance).

Sim 2: projectile launcher

Set the launch speed, angle and height. Strobe dots (ghosts) every 0.25 s show equal sideways steps. Green arrows show vx and vy. Predict the vy-t graph first.

Things to try: (1) 45° gives the longest range on level ground. (2) 30° and 60° give the same range. (3) At the top, vy = 0 but vx is unchanged and a is still −9.8 m/s². (4) Set angle 0° and height 20 m: that is the "thrown sideways" ball from the story.

Worked examples

Dropping a stone basic

A stone is dropped from rest from a bridge 20 m above the water. How long does it fall and how fast is it going when it hits?

Show solution
Show all steps as text

Up +, v0 = 0, a = −9.8 m/s², Δy = −20 m.

−20 = ½(−9.8)t² → t = √(40/9.8) = 2.02 s.

v = −9.8 × 2.02 = −19.8 m/s, so 19.8 m/s downward. Check: v² = 2(9.8)(20) = 392, √392 = 19.8 ✓.

Straight up and back basic

A ball is thrown straight up at 15 m/s. Find (a) the time to reach the top, (b) the maximum height above the hand, (c) the velocity when it comes back to the hand.

Show solution
Show all steps as text

(a) 0 = 15 − 9.8t → t = 1.53 s.

(b) 0 = 15² − 2(9.8)h → h = 225/19.6 = 11.5 m.

(c) By symmetry it returns at 15 m/s downward (v = −15 m/s) after 3.06 s.

Rolling off a table medium

A marble rolls off a 1.25 m high table at 3.0 m/s. How far from the table edge does it land?

Show solution
Show all steps as text

y: −1.25 = −4.9t² → t = √(1.25/4.9) = √0.255 = 0.505 s.

x: Δx = 3.0 × 0.505 = 1.52 m.

Notice: the fall time depends only on the height. A faster marble lands farther out but at the same time.

A soccer kick medium

A ball is kicked from level ground at 25 m/s, 53° above horizontal. Find its maximum height, time of flight and range.

Show solution
Show all steps as text

v0x = 25 cos 53° = 15.0 m/s; v0y = 25 sin 53° = 20.0 m/s.

Max height: h = v0y²/(2g) = 20.0²/19.6 = 20.4 m.

Time of flight: t = 2v0y/g = 2(20.0)/9.8 = 4.08 s.

Range: 15.0 × 4.08 = 61.2 m.

Thrown upward at an angle from a cliff AP

A ball is thrown from the edge of a 30 m cliff at 20 m/s, 30° above horizontal. How long is it in the air and how far from the base of the cliff does it land?

Show solution
Show all steps as text

v0x = 17.3 m/s, v0y = 10.0 m/s. Take the cliff top as y = 0, so landing is at y = −30 m.

−30 = 10.0t − 4.9t² → 4.9t² − 10.0t − 30 = 0.

t = [10.0 + √(100 + 4·4.9·30)] / (2·4.9) = (10.0 + √688)/9.8 = (10.0 + 26.2)/9.8 = 3.70 s (the negative root is before the throw).

Δx = 17.3 × 3.70 = 64.0 m.

AP tip: you could also split it: up to the top (1.02 s, rises 5.10 m), then fall 35.1 m from rest (√(2·35.1/9.8) = 2.68 s). Total 3.70 s ✓.

Practice (AP style)

  1. A ball is thrown straight up. Ignore air resistance. At the very top of its path, what are its velocity and acceleration?

    Show answer

    (B). Gravity never switches off. If a were zero at the top, v would stay zero and the ball would hover.

  2. From the same height, ball X is dropped and ball Y is thrown horizontally at the same instant. Ignore air resistance. Which hits the level ground first?

    Show answer

    (C). Both start with vy = 0 and have ay = −9.8 m/s², so their vertical motions are identical.

  3. A ball is thrown horizontally at 8.0 m/s from a height of 4.9 m. How far from the launch point (horizontally) does it land?

    Show answer

    (B). 4.9 = 4.9t² → t = 1.0 s. Δx = 8.0 × 1.0 = 8.0 m.

  4. A ball is thrown straight up at 19.6 m/s. How long until it returns to the launch height?

    Show answer

    (C). Up: 19.6/9.8 = 2.0 s. Down takes the same 2.0 s. Total 4.0 s.

  5. A ball is launched at 20 m/s, 30° above horizontal. What is its speed at the top of its path?

    Show answer

    (C). At the top only vy is zero. vx = 20 cos 30° = 17.3 m/s all flight long.

  6. On level ground with no air resistance, a ball launched at 25° above horizontal has the same range as a ball launched at the same speed at which angle?

    Show answer

    (B). Complementary angles (adding to 90°) give equal ranges: range = v0² sin(2θ)/g and sin 50° = sin 130°. The 65° ball goes higher and stays up longer, but moves slower sideways.

  7. Up is positive. Which best describes the graph of vy against time for a projectile launched upward at an angle, from launch until it lands at the same height?

    Show answer

    (B). Constant acceleration means a straight v-t line with slope = a = −9.8 m/s². (A) is the graph of speed in y, not velocity. (D) is the y-t graph.

  8. Short free response (experimental design). A student rolls a ball off a table of height h at different speeds v (measured with a photogate at the edge) and measures the landing distance d from the table.

    (a) What graph should the student plot to get a straight line? (b) What does the slope mean physically? (c) If the slope is 0.45 s, what is the table height?

    Show answer

    (a) Plot d (vertical axis) against v (horizontal axis). Since d = v t and the fall time t is the same for every run, d is proportional to v.

    (b) The slope is the fall time t = √(2h/g).

    (c) h = ½ g t² = ½ (9.8)(0.45)² = 0.99 m.

    Scoring idea: 1 point for d vs v, 1 point for "slope = fall time, same for all trials", 1 point for the height with units.

AP question types for this topic

Four short free response questions, one of each AP Physics 1 free response type, all about this topic. Write your answer first, then open the worked answer and score yourself with the points shown.

1. Mathematical Routines

Lucia's marble rolls off a table of height h with horizontal speed v0. Ignore air resistance.

(a) Derive expressions for the time to land and the horizontal distance in terms of h, v0 and g.

(b) Calculate both for h = 1.2 m and v0 = 2.5 m/s.

(c) Find the speed and direction of the marble just before it lands.

Show worked answer and scoring
  1. (a) Vertically from rest: h = ½gt2, so t = √(2h/g). Horizontally the speed stays v0: x = v0√(2h/g). 1 point for the time, 1 point for the distance.
  2. (b) t = √(2.4 ÷ 9.8) ≈ 0.49 s; x = 2.5 × 0.495 ≈ 1.24 m. 1 point.
  3. (c) vy = gt = 9.8 × 0.495 ≈ 4.85 m/s down. Speed = √(2.52 + 4.852) ≈ 5.5 m/s, at tan−1(4.85/2.5) ≈ 63° below horizontal. 1 point for the speed, 1 point for the angle.

Total: 5 points.

2. Translation Between Representations

Kenji kicks a ball from level ground at 12 m/s, 30° above horizontal. It flies up and lands back on the ground. Take up as positive and ignore air resistance.

vt
A
vt
B
vt
C
vt
D

Velocity–time graphs (time axis across the middle). A: a straight line falling from above the axis to below it. B: a flat line above the axis. C: a V shape. D: an upside-down U.

(a) Which graph shows the vertical velocity? Justify, and find when it crosses zero.

(b) Which graph shows the horizontal velocity? Justify.

(c) Graph D has the same shape as the ball's path. Explain why it is wrong for either velocity, and describe the vertical acceleration–time graph.

Show worked answer and scoring
  1. (a) A. vy starts at 12 sin 30° = 6.0 m/s and changes by −9.8 m/s each second: a straight line of slope −9.8. Zero at the top, t = 6.0 ÷ 9.8 ≈ 0.61 s, ending at −6.0 m/s. 1 point for A, 1 point for 0.61 s.
  2. (b) B. No horizontal force, so vx = 12 cos 30° ≈ 10.4 m/s the whole time. 1 point.
  3. (c) D copies the path, which is height against distance, not velocity against time. The ay–t graph is a flat line at −9.8 m/s2 for the whole flight. 1 point for the explanation, 1 point for the flat line.

Total: 5 points.

3. Experimental Design and Analysis

Yusuf wants to measure g. He has a steel ball held by an electromagnet, an electronic timer that starts when the magnet releases the ball and stops when it hits a switch pad, and a metre stick.

(a) Describe his procedure.

(b) Derive what he should plot to get a straight line, and what the slope means.

(c) Use his data to find g.

h (m)0.200.400.600.80
t (s)0.2030.2850.3510.403

Show worked answer and scoring
  1. (a) Measure the drop height from the bottom of the ball to the pad. Release and record the time; repeat three times at each of several heights and average. 1 point for measuring height and time, 1 point for several heights with repeats.
  2. (b) From rest h = ½gt2. Plot h against t2: a straight line through the origin, slope = g/2, so g = 2 × slope. 1 point for the equation, 1 point for the plot and slope.
  3. (c) t2 = 0.041, 0.081, 0.123, 0.162 s2. Best-fit slope ≈ 4.93 m/s2, so g ≈ 9.9 m/s2, close to 9.8. 1 point for the t² values, 1 point for g.

Total: 6 points.

4. Qualitative/Quantitative Translation

Two marbles roll off the same table, 0.90 m high. Marble P leaves at 1.5 m/s, marble Q at 3.0 m/s. Mateo claims: “Q lands twice as far away, but both land at the same moment.”

(a) Explain in words whether Mateo is right.

(b) Derive the fall time and landing distance, and calculate them for each marble.

(c) Connect your expressions in (b) to your reasoning in (a).

Show worked answer and scoring
  1. (a) Yes. Vertical and horizontal motions are independent. Both start with zero vertical velocity from the same height, so they fall for the same time; in that time Q, moving sideways twice as fast, goes twice as far. 1 point for same time, 1 point for twice the distance.
  2. (b) t = √(2h/g) = √(1.8 ÷ 9.8) ≈ 0.43 s for both. x = v0t: P ≈ 0.64 m, Q ≈ 1.29 m. 1 point for the time, 1 point for both distances.
  3. (c) The time expression has only h and g, no v0, so sideways speed cannot change the fall time; x = v0t is proportional to v0. That is the maths form of (a). 1 point.

Total: 5 points.

Common mistakes

The mistake: "At the top, v = 0, so a = 0."

Why it is wrong: Acceleration is how fast velocity changes. At the top, v is changing from up to down, so a = −9.8 m/s².

How to spot it: If a were 0 at the top, the ball would stay there forever.

The mistake: Letting the horizontal velocity slow down during the flight.

Why it is wrong: Gravity points straight down. With no air resistance, nothing pushes sideways, so vx is constant.

How to spot it: Strobe dots of a projectile are equally spaced sideways. If yours bunch up, something is off.

The mistake: Using the full launch speed v0 in the x or y equation.

Why it is wrong: Each direction gets only its own part: v0 cos θ for x, v0 sin θ for y (θ measured from horizontal).

How to spot it: Every equation you write should have an x or y label on each symbol.

The mistake: Sign errors: writing a = +9.8 while also calling up positive.

Why it is wrong: Gravity points down. If up is +, a = −9.8 m/s². A dropped ball would otherwise fly upward in your maths.

How to spot it: State your positive direction at the top of every problem. Check that a falling object has negative v.

The mistake: "A heavier ball falls faster."

Why it is wrong: Without air resistance, all objects fall with the same acceleration. Mass does not appear in any kinematics equation.

How to spot it: If your answer depends on mass in a free-fall question, recheck. (Air resistance is a Unit 2 story.)