1. Read for the knowns, then the unknown

Before any equation, turn the sentence into a list. Every number in the problem gets a symbol, a value and a unit. Then write the single quantity you are asked for. Half of all wrong answers come from solving for the wrong thing, and the list makes that impossible. Include what the words imply: "starts from rest" means v₀ = 0; "just before it hits the ground" means the height is zero; "constant speed" means the net force is zero.

2. Draw it

A sketch with the directions marked is not a courtesy to the marker; it is where you decide which direction is positive, and that decision has to hold for every quantity that follows. Forces get a free-body diagram. Motion in two dimensions gets split into components before anything else happens. Circuits get labelled currents with arrows.

3. Pick the principle, then the formula

Ask which idea connects your knowns to your unknown: conservation of energy, Newton's second law, a kinematics relation, a circuit law. Only then pick the formula, and check its conditions against the problem before you use it. The formula sheet on this site lists those conditions for every entry, which is the part most sheets leave out.

4. Do the algebra with the units attached

Rearrange symbolically first, substitute numbers last. Carry the units through every line: a unit that does not cancel to what you expect is the fastest possible signal that the wrong formula, or the wrong quantity, went in. Keep signs consistent with the direction you chose in step 2.

5. Check the answer before you believe it

Sign: does a negative make sense here? Size: is a car really going 400 m/s? Limits: if a quantity went to zero or to infinity, would the formula still behave? Two seconds of checking catches most of what the method missed.

The method on a real problem

Problem. A ball is launched from level ground at 20 m/s, 30° above the horizontal. Find its time of flight, range and peak height.

  1. Split the launch velocity: v₀ₓ = 20 cos 30° = 17.3 m/s and v₀ᵧ = 20 sin 30° = 10.0 m/s.
  2. Time of flight on level ground: t = 2v₀ᵧ / g = 20.0 / 9.8 = 2.04 s.
  3. Range: R = v₀ₓt = (17.3)(2.04) = 35.3 m, which matches R = v₀² sin 2θ₀ / g = 400(0.866) / 9.8 = 35.3 m.
  4. Peak height: h = v₀ᵧ² / 2g = 100 / 19.6 = 5.1 m.

Answer. It stays in the air 2.0 s, lands about 35 m away, and peaks 5.1 m above the ground. The full guide to the formulas used is at Projectile motion formulas.

Questions people ask

Why do I understand the lecture but freeze on the problem?

Because understanding a principle and choosing one under pressure are different skills, and only the second is practiced by doing problems. The method on this page makes the choice explicit: list what you know, name the principle, then the formula. Freezing usually happens at the naming step, and naming gets faster with repetition.

Should I memorize the formulas?

Memorize the conditions more than the symbols. A formula sheet gives you the symbols; nothing on the sheet tells you that the kinematics equations need constant acceleration or that the ideal gas law wants kelvin. The formula pages on this site pair every formula with its conditions for exactly that reason.

How many problems should I do?

Fewer than you think, done properly: every step written, units carried, answer checked for sign and size. Ten problems with the method beat forty done by pattern-matching to the previous one.