Kinematics

Problem-solving process

01 / Problem-solving process

Work each problem before opening the solution

Pause before each solution. Write the knowns, select an equation, and complete the problem before using the worked solution as a check.

Assumptions

Except for the explicit air-resistance comparison, each problem uses one-dimensional vertical motion with gravity as the only retained interaction.
For these near-Earth classroom-scale height changes, treat g as approximately constant.
Choose a positive direction before assigning signs and keep that convention for the entire problem.

Read and identify the target
Read the complete question. Name what you must find before selecting an equation.
List stated and hidden knowns
time, displacement, initial velocity, final velocity, acceleration
Phrases such as falls from rest or highest point supply values that may not be printed beside a number.
Declare the sign convention
Decide whether upward or downward is positive. Then assign the signs of displacement, velocity, and g from that decision.
Select and rearrange
Choose an equation containing the knowns and requested unknown. Isolate the target before substituting when that makes the algebra clearer.
Substitute, solve, and check
Carry signs and units into the calculation. Check the unit, direction, magnitude, and physical meaning of the result.

Result: Practice is the point of this lesson. Use each worked solution to inspect your reasoning, not as a replacement for doing the problem.

02 / Choosing a value of g

AP Physics Focus: choose the value of g deliberately

For AP Physics work, using |g| = 10 m/s² is allowed and encouraged unless a problem supplies another value or specifically requires greater precision.

The worked examples below retain 9.81 m/s² because that is the value used in the source lesson. Whichever value the problem calls for, assign its sign from the chosen positive direction and use the same value consistently throughout the solution.

03 / Falling-object problems

Problem 1: A chunk of ice falls from a glacier

A chunk of ice falls freely through 30.0 m before hitting the water. How long does the fall take?

Assumptions

Choose downward as negative, with the release point at zero and the water below it.
The ice begins from rest, so = 0.
Gravity is the only retained interaction and a = g = −9.81 m/s².

List the signed knowns
= −30.0 m; v(initial) = 0 m/s; a = −9.81 m/s²
The displacement and acceleration are both negative under the chosen convention. The impact velocity is unknown; hitting the water does not make it zero.
Choose the displacement-time equation
= v(initial)t + ½at²
This equation contains the three known quantities and the requested time while omitting the unknown final velocity.
Use the zero initial velocity and rearrange
= ½at²; t = √(2/a)
The initial-velocity term disappears. The two negative signed quantities will make the ratio under the square root positive.
Substitute with units
t = √[2(−30.0 m)/(−9.81 m/s²)]
Meters divided by meters per second squared gives seconds squared; the square root gives seconds.
Solve and interpret
t = 2.47 s
Time is a positive interval. The glacier-to-water sign model is release at zero, water at −30.0 m, initial velocity zero, and g directed downward.

Result: The chunk of ice falls for 2.47 s before reaching the water.

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Problem 2: The rock's acceleration is the same while rising, at the top, and while falling
Stage with upward chosen positive	Velocity	Acceleration	Reasoning
Rising	Upward and positive, with decreasing magnitude	a = g = −9.81 m/s²	Downward gravity slows the upward-moving rock
Highest point	v = 0 for one instant	a = g = −9.81 m/s², not zero	If acceleration stayed zero, the rock would hover instead of beginning to fall
Falling	Downward and negative, with increasing magnitude	a = g = −9.81 m/s²	The same downward acceleration continues after the velocity reverses

The three stages use the same upward-positive coordinate system. Velocity changes from positive to zero to negative, but acceleration remains the same nonzero downward vector. The acceleration arrows would have equal direction and magnitude at every stage; labels and values carry this meaning without color.

04 / Direction and air resistance

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Problem 3: Separate the velocity change from the sign of gravity
Question	Answer	Upward-positive sign model
When is velocity zero?	At the highest point, for one instant	v changes from positive to zero
Does velocity change direction?	Yes, after the highest point	v becomes negative during the descent
Does gravity change sign?	No, not while the coordinate convention remains fixed	g remains −9.81 m/s² on the way up, at the top, and on the way down

Velocity and acceleration answer different questions. Under one upward-positive convention, the object's velocity reverses direction, but the downward acceleration due to gravity keeps the same negative sign throughout the motion.

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Problem 4: Returning to the launch height with and without air resistance
Model	Maximum height	Return motion at the starting height
Negligible air resistance	The ideal free-fall peak	Return speed equals launch speed in magnitude, with the velocity directed downward
Non-negligible air resistance	Lower than the ideal free-fall peak	Return speed is smaller in magnitude than the launch speed

Air resistance opposes the motion and breaks the ideal up-and-down symmetry. With drag, the object reaches a lower peak and returns to the launch height more slowly. This comparison is qualitative: no drag coefficient, terminal-speed model, or numerical result is introduced.

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Problem 5: A ball launched upward at 15.0 m/s from y₀ = 0, with upward positive and g = −9.81 m/s²
Time	Displacement from release	Velocity	Physical interpretation
0.500 s	= (15.0 m/s)(0.500 s) + ½(−9.81 m/s²)(0.500 s)² = +6.27 m	= 15.0 m/s + (−9.81 m/s²)(0.500 s) = +10.1 m/s	Above the release point and still rising
1.00 s	= (15.0 m/s)(1.00 s) + ½(−9.81 m/s²)(1.00 s)² = +10.1 m	= 15.0 m/s + (−9.81 m/s²)(1.00 s) = +5.19 m/s	Higher and still rising, more slowly
1.50 s	Board/calculator display: = (15.0 m/s)(1.50 s) + ½(−9.81 m/s²)(1.50 s)² = +11.46 m; reported to three significant figures: +11.5 m	= 15.0 m/s + (−9.81 m/s²)(1.50 s) = +0.285 m/s	Near the highest point but still rising
2.00 s	Board/calculator display: = (15.0 m/s)(2.00 s) + ½(−9.81 m/s²)(2.00 s)² = +10.38 m; reported to three significant figures: +10.4 m	= 15.0 m/s + (−9.81 m/s²)(2.00 s) = −4.62 m/s	Still above the release point but now moving downward

Each row uses = t + ½at² and = + at. The calculator displacement values are +6.27 m, +10.1 m, +11.46 m, and +10.38 m; the last two are reported to a consistent three significant figures as +11.5 m and +10.4 m. The displacement first increases, nearly reaches its maximum by 1.50 s, and then decreases by 2.00 s. The velocity remains positive through 1.50 s and is negative at 2.00 s. The +0.285 m/s entry is final velocity at 1.50 s, not average velocity. Signed values and direction words carry the full snapshot meaning without relying on arrow color or length.

05 / Upward-launch problems

Problem 6: Minimum jump speed

What upward launch velocity is required for a basketball player to rise 1.25 m?

Assumptions

Choose upward as positive, so = +1.25 m and a = g = −9.81 m/s².
For the minimum launch speed, the player reaches the required height at the highest point, where = 0.
The printed and board-verified rise is 1.25 m; the isolated 1.5 m caption rendering is not used.

List the knowns and unknown
= +1.25 m; v(final) = 0 m/s; a = −9.81 m/s²; v(initial) = ?
Zero velocity at the highest point supplies the hidden known. The launch velocity is the target.
Choose the no-time equation
v(final)² = v(initial)² + 2a
This equation contains displacement, acceleration, both endpoint velocities, and no time.
Rearrange for the initial velocity
v(initial) = √(−2a)
With = 0, move the acceleration-displacement term to the other side before taking the square root.
Substitute
v(initial) = √[−2(−9.81 m/s²)(1.25 m)]
Acceleration times displacement gives square meters per square second, so the square root gives meters per second.
Select the physically correct root
v(initial) = +4.95 m/s
The squared equation permits two algebraic signs, but this launch is upward under an upward-positive convention, so choose the positive root.

Result: The minimum upward launch velocity is +4.95 m/s.

Problem 7: Dolphin height and time

A dolphin leaves the water moving upward at 13.0 m/s. Find its maximum rise above the water and its total time in the air.

Assumptions

Choose upward as positive, so = +13.0 m/s and a = g = −9.81 m/s².
At maximum height, = 0.
The total-airtime completion assumes the dolphin returns to the same water level with negligible air resistance and approximately constant g.

Choose the no-time equation for maximum height
v(final)² = v(initial)² + 2a
The peak supplies = 0, and the equation contains the launch velocity, acceleration, and vertical displacement.
Rearrange and solve for the rise
= −v(initial)²/(2a) = −(13.0 m/s)²/[2(−9.81 m/s²)] = 8.61 m
This matches the verified maximum-height result from the lesson.
Calculate the time to the top
a = /t; t(up) = [0 − 13.0 m/s]/(−9.81 m/s²) = 1.33 s
The 1.33 s calculation runs from launch to the highest point. It is ascent time, not total time in the air.
Editorial physics completion for the printed total-airtime request
t(total) = 2t(up) = 2(1.325 s) = 2.65 s
The source solution stops after finding the ascent time. Doubling it completes the printed total-airtime request and is valid only under the stated same-level, negligible-air-resistance, approximately constant-g symmetry.
Keep the two results distinct
t(up) = 1.33 s; t(total) = 2.65 s
Labeling 1.33 s as total airtime would leave the printed question unanswered and would conflict with the physics of the symmetric ideal model.

Result: The maximum rise is 8.61 m and the ascent time is 1.33 s. Under the explicit same-level, no-drag, constant-g assumptions, the completed total airtime is 2.65 s. Keep ascent time and total airtime labeled as different intervals.

Three more challenges from the classroom lesson

These short problems target equation choice and interpretation rather than copying a long worked solution. Write the knowns and chosen positive direction before opening each answer.

Use the value of g stated in each prompt. The first two use the AP-friendly approximation |g| = 10.0 m/s²; the vacuum-tube calculation uses 9.81 m/s².

Check your understanding

An object is thrown straight upward from ground level at 30 m/s and returns to ground level. Neglect air resistance and use |g| = 10.0 m/s². About how long is the object in the air?

Show the answer

6.0 s.

The upward trip takes 30 m/s ÷ 10.0 m/s² = 3.0 s. Under the same-height, negligible-drag model, the downward trip takes the same time, so the total airtime is 6.0 s.

Check your understanding

An object is dropped from rest. Use |g| = 10.0 m/s². Compare the magnitude of its displacement from the release point after 1.0 s with its displacement from the release point after 3.0 s.

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After 1.0 s the displacement magnitude is 5.0 m. After 3.0 s it is 45 m, which is nine times as large.

For an object dropped from rest, displacement magnitude is ½|g|t². At 1.0 s: ½(10.0)(1.0)² = 5.0 m. At 3.0 s: ½(10.0)(3.0)² = 45 m. The total displacement grows with t², not directly with t.

Check your understanding

A coin is dropped from rest in a vacuum tube. Using |g| = 9.81 m/s², how far does it fall in 0.30 s?

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0.44 m downward.

The displacement magnitude is ½|g|t² = ½(9.81 m/s²)(0.30 s)² = 0.441 m, which rounds to 0.44 m. If upward is positive, the signed displacement is −0.44 m.

06 / Final check

Check your understanding

What should you carry into the next vertical-motion problem?

Show the answer

Infer hidden values, declare and keep a positive direction, assign the sign of g from that convention, distinguish velocity from acceleration, choose an equation from the knowns and target, and check units and physical meaning.

A released object may have = 0, while an object at its highest point has v = 0 for only an instant. Gravity's sign does not change merely because velocity reverses. Displacement and velocity together tell you where an object is and which way it is moving. When a printed request and a worked calculation cover different intervals, name the interval honestly before completing the physics. This summary adds no eighth problem and imports no content from the following isolating-variables lesson.

Continue the sequence

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This page follows the Free Fall Example Problems lesson in the Kinematics course sequence.

See the Kinematics course