Motion in plane

What you will understand

Resolve a projectile into components with one shared clock, explain its nonzero speed at the top and choose flight time from the actual landing height. Then connect the vector method to relative velocity and circular motion.

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Quick answer: keep the clock, separate the components

Choose an inertial ground frame, horizontal $+x$ and upward $+y$. For a projectile with no air resistance and constant downward $g$, $a_x=0$ and $a_y=-g$. Use $x=x_0+u_xt$ and $y=y_0+u_yt-\tfrac12gt^2$ with the same time $t$. Also $v_x=u_x$ and $v_y=u_y-gt$.

The vertical equation determines when the projectile reaches the specified landing height. The horizontal equation then determines where it lands. Horizontal and vertical velocities need not become zero together.

See what changes at the top of the trajectory

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State the conditions before using a shortcut

For launch speed $u$ at angle $\theta$ above horizontal, $u_x=u\cos\theta$ and $u_y=u\sin\theta$. With upward launch, equal launch and landing heights, constant $g$ and no drag, flight time is $2u_y/g$ and horizontal range is $u^2\sin(2\theta)/g$. The familiar 45-degree maximum range also assumes fixed speed and equal endpoint heights. If any condition changes, return to the component equations.

Change the landing height, then solve for time

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Use vectors beyond projectile motion

Relative velocity needs frame labels: $\vec v_{A/B}=\vec v_{A/G}-\vec v_{B/G}$, with both right-hand vectors measured in the same ground frame. For example, $(6\hat i+2\hat j)-(2\hat i-\hat j)=4\hat i+3\hat j$ in metres per second; the relative speed is 5 m/s. Subtract components before taking a magnitude.

In uniform circular motion, speed is constant but the velocity direction changes. Acceleration points inward with magnitude $v^2/r$; it is not zero. A 4 m/s speed on a 2 m radius gives 8 m/s² inward. These are different acceleration models from the constant-vector projectile example.

Common mistakes to catch

Do not set the whole velocity to zero at maximum height. Do not remove gravity at the top or treat horizontal velocity as horizontal acceleration. Distance along a curved trajectory is not its horizontal range. Equal heights on the ascending and descending parts give opposite vertical-velocity signs, not identical velocity vectors. A constant speed on a curved path does not mean zero acceleration.

Try it yourself 1: velocity and acceleration at the top

At the top of the first trajectory, which combination is correct? A. Speed 0, acceleration 0. B. Speed 16 m/s, acceleration 10 m/s² downward. C. Speed 20 m/s, acceleration 0.

Try it yourself 2: choose the landing event

A horizontal launch at 10 m/s starts 20 m above ground, with $g=10\,\mathrm{m\,s^{-2}}$. Which ground-landing result is correct? A. 0 s and 0 m; B. 2.0 s and 20 m; C. 4.0 s and 40 m.

A useful check on your next visit

Redraw the velocity and acceleration at the top, then explain why a horizontally launched object still takes time to reach lower ground. Before your next question, name the axes, common clock and landing height. Use the specific missing distinction to choose the next practice item.

Choose your next chapter task

Explain the distinction you missed in one sentence, then use it in a fresh chapter question.

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Sources and review

Technical review: Codex (automated). Primary textbook sources checked on 23 October 2026. Diagrams, worked numerical examples and self-checks are original teaching material. Human subject review is not claimed.