Figluma

Learn exam mechanics by writing the next equation, not by reading a wall of notes.

5 live modelsGuided problem solvingDirectory Validation

Start hereWork checkpoint 1: Choose projectile axes and origin.

Write one line in Scratch working, then press Check this line before revealing.

What it doesTurns one physics problem into small equation decisions.

Use it when the hard part is choosing the force, component, sign, or equation.

Learning loopDiagram, equation, feedback, next checkpoint.

Do not read everything first. Move through the rail and explain each equation.

1 Pick a problem

Mechanics

Block on a rough inclineForces / Inclined planes / Friction Crate pulled across a rough floorForces / Horizontal motion / Friction Table block and hanging massConnected particles / Friction / Newton's laws Passenger in an accelerating elevatorNewton's laws / Apparent weight / Vertical motion Projectile launched from a cliffProjectile motion / Kinematics / Vectors

A ball is launched from a 20 m high cliff with speed 18 m s^-1 at 35 degrees above the horizontal. Ignore air resistance. Find the time of flight, horizontal range from the base of the cliff, and impact speed just before it hits the ground.

Your job: connect the known values to the unknown by building the equation one checkpoint at a time.

Knowns and target

Givens

$v_{0}$ Launch speed: $18 m s^{- 1}$
$θ$ Launch angle above horizontal: $3 5^{\circ}$
$y_{0}$ Launch height: $20 m$
$g$ Gravitational field strength: $9.8 m s^{- 2}$

Unknowns

$t$ Time of flight: $? s$
$R$ Horizontal range from cliff base: $? m$
$∣ v ∣$ Impact speed: $? m s^{- 1}$

+x is horizontal to the right. +y is upward, with launch point y = 20 m and ground at y = 0.

Assumptions

Air resistance is negligible, so horizontal acceleration is zero.
The only vertical acceleration is gravity, a_y = -g.
The launch speed and angle are measured at the instant the ball leaves the cliff.
The impact point is on level ground 20 m below the launch point.

Constraints

Horizontal and vertical motion share the same time t but have different acceleration.
Horizontal velocity is constant: v_x = v_0 cos(theta).
Vertical position uses the signed displacement to the ground: 0 = y_0 + v_0 sin(theta)t - 1/2 gt^2.
Impact speed must combine horizontal and vertical velocity components.

Projectile launched from a cliff

2 Try this checkpoint before opening the worked equation.

Look atSetupChoose projectile axes and origin

Equation jobChoose projectile axes and originUse independent horizontal and vertical axes. The launch point starts above the ground, so the vertical position equation must carry y_0.

Watch forCommon mistakeMaking gravity positive in the +y upward convention.

Equation lens

+ x

right; +y = upward; y_0 = 20 m

Choose projectile axes and originBefore revealing the worked equation, decide what "Choose projectile axes and origin" must establish from the diagram and givens.

Hint before equation

Set the axes, origin, sign convention, or linked-motion choice before writing algebra.

Making gravity positive in the +y upward convention.

Equation-choice checkWhat feature of the diagram, sign convention, or givens makes "Choose projectile axes and origin" the right next equation?

Step 1 of 6SetupChoose projectile axes and origin

Use the diagram focus, problem state, and scratch workspace first.

3 Checkpoint path

Follow one checkpoint at a time. Hidden equations are not skipped; they are the task.

Choose projectile axes and origin
Not yet
Set the axes, origin, sign convention, or linked-motion choice before writing algebra.
Watch: Making gravity positive in the +y upward convention.
Resolve the launch velocity
Not yet
Resolve only onto the chosen axes and check which part of each vector belongs in this step.
Watch: Using the full launch speed as the horizontal velocity instead of v_0 cos(theta).
Write the vertical position equation
Not yet
Separate driving terms from opposing terms, then assign signs from the positive direction.
Watch: Using the same-height range formula v_0^2 sin(2theta) / g even though the projectile lands below launch height.
Solve the quadratic for flight time
Not yet
Substitute after the model equation is settled, then check units and the requested interpretation.
Watch: Reporting the vertical impact speed alone instead of combining v_x and v_y.
Use horizontal motion for range
Not yet
Substitute after the model equation is settled, then check units and the requested interpretation.
Watch: Reporting the vertical impact speed alone instead of combining v_x and v_y.
Combine impact velocity components
Not yet
Substitute after the model equation is settled, then check units and the requested interpretation.
Watch: Reporting the vertical impact speed alone instead of combining v_x and v_y.

Common traps

Using the full launch speed as the horizontal velocity instead of v_0 cos(theta).
Using the same-height range formula v_0^2 sin(2theta) / g even though the projectile lands below launch height.
Solving only for time to the top of the path, v_0 sin(theta) / g, instead of time to the ground.
Dropping the initial height term from the vertical displacement equation.
Making gravity positive in the +y upward convention.
Reporting the vertical impact speed alone instead of combining v_x and v_y.

Your workspace

Do this nowWrite the next line for: Choose projectile axes and origin

Use the diagram cue, explain why the equation fits, then check one scratch line. Reveal only after an attempt.

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Solve checklist0 / 4 marked

SetupAxes, sign convention, model constraints, and linked-motion/origin choices are stated.
Choose projectile axes and origin
1 common trap tied to this row.
ComponentsResolved components, force directions, normal/friction setup, or velocity split are correct.
Resolve the launch velocity
1 common trap tied to this row.
Net-force / governing equationThe main Newton's law or motion equation uses the right model, signs, and shared variables.
Write the vertical position equation
3 common traps tied to this row.
ResultThe final rearrangement, numeric value, units, and direction/speed interpretation are correct.
Solve the quadratic for flight timeUse horizontal motion for rangeCombine impact velocity components
1 common trap tied to this row.

Why does this equation fit?

One sentence is enough. The aim is to connect the line to the diagram, not to write notes.

Write one line to check