Horizontal and Vertical Equations of Motion to solve problems

From subtopic 8.1, a projectile launched with speed $V$ at angle $\theta$ (up positive, $g = 10$) obeys:

$x = V\cos\theta\,t \qquad\qquad y = V\sin\theta\,t - \tfrac12 g t^2$

The horizontal one is just distance = speed × time. The vertical one is the SUVAT equation $s = ut + \tfrac12 at^2$ with $u = V\sin\theta$ and $a = -g$.

The full SUVAT toolkit (vertical direction). You will often need other SUVAT equations vertically. With $u = V\sin\theta$ and $a = -g$:

• $v = u + at$ gives the vertical velocity: $v_y = V\sin\theta - gt$.
• $v^2 = u^2 + 2as$ links vertical velocity and height without time.
• $s = ut + \tfrac12 at^2$ gives height from time (the equation above).

How to choose. Ask: which vertical quantities do I know and want? Then pick the SUVAT equation containing exactly those. If a height is given and you want the velocity there, use $v^2 = u^2 + 2as$ (no $t$). If you know the time, use $v = u + at$.

Cambridge tip. Keep $g = 10$ symbolic until the last step if it tidies the algebra, but it is usually fine to substitute $g = 10$ immediately — Cambridge expects it.

On level (horizontal) ground, the projectile lands when its height returns to zero, i.e. $y = 0$ again.

Time of flight. Set $y = 0$: $V\sin\theta\,t - \tfrac12 g t^2 = 0 \;\Rightarrow\; t\left(V\sin\theta - \tfrac12 g t\right) = 0$ The solution $t = 0$ is the launch; the landing is at $T = \frac{2V\sin\theta}{g}$

Range. Substitute $T$ into $x = V\cos\theta\,t$: $R = V\cos\theta \cdot \frac{2V\sin\theta}{g} = \frac{2V^2\sin\theta\cos\theta}{g} = \frac{V^2\sin 2\theta}{g}$ using the identity $2\sin\theta\cos\theta = \sin 2\theta$.

Maximum range. Since $\sin 2\theta \leq 1$, the range is greatest when $\sin 2\theta = 1$, i.e. $2\theta = 90^\circ$, so $\theta = 45^\circ$. The maximum range is then $\dfrac{V^2}{g}$.

Cambridge tip. These $R$ and $T$ formulae are quick to quote, but Paper 3 often wants you to derive them, or they apply only on level ground. If the projectile lands at a different height, do not use them — go back to $y = 0$ (or $y =$ that height) directly.

At the highest point of the path, the projectile is moving purely horizontally — its vertical velocity is momentarily zero. This is the key that unlocks the greatest height.

Method 1 — using $v^2 = u^2 + 2as$ (no time needed). Vertically, $u = V\sin\theta$, $v = 0$ at the top, $a = -g$, $s = H$: $0 = (V\sin\theta)^2 - 2gH \;\Rightarrow\; H = \frac{V^2\sin^2\theta}{2g}$

Method 2 — find the time to the top, then the height. Vertical velocity $v_y = V\sin\theta - gt = 0$ at the top gives $t = \dfrac{V\sin\theta}{g}$ (exactly half the time of flight). Substitute into $y$: $H = V\sin\theta \cdot \frac{V\sin\theta}{g} - \tfrac12 g\left(\frac{V\sin\theta}{g}\right)^2 = \frac{V^2\sin^2\theta}{2g}$

Both routes agree. Method 1 is faster when you only want the height; Method 2 is useful when you also want the time.

Projectile launched from a height. If the projectile starts at height $h$ above where it lands (e.g. off a cliff), the greatest height above the launch point is still $\dfrac{V^2\sin^2\theta}{2g}$, but the time of flight and range are longer — you must set $y = -h$ (with up positive, origin at launch) and solve the quadratic for $t$, taking the positive root.

Cambridge tip. "Greatest height" almost always means above the point of projection, unless the question explicitly says "above the ground". Read carefully when the launch is from a cliff or building.

To find the projectile's velocity at a moment in its flight, get both components and recombine them.

At a given time $t$:

• Horizontal: $v_x = V\cos\theta$ (unchanged).
• Vertical: $v_y = V\sin\theta - gt$ (use $v = u + at$). It is positive going up, negative coming down.

At a given height $y$ (when time isn't known): use $v_y^2 = (V\sin\theta)^2 - 2gy$ to get the vertical component directly, then take $\pm$ as appropriate (the projectile passes each height twice — once up, once down).

Recombine the components: $\text{speed } v = \sqrt{v_x^2 + v_y^2}, \qquad \tan\phi = \frac{v_y}{v_x}$ where $\phi$ is the angle of the velocity to the horizontal. A negative $v_y$ means the projectile is moving downward at that instant.

Watch the sign of $v_y$. Because $v_y$ changes sign, you frequently get two answers for "find the time when the speed is …" or "the direction is …" — one on the way up, one on the way down. State both unless the question restricts you.

Cambridge tip. Always quote the direction with words ("above" / "below the horizontal") or with a correctly signed angle. A bare number like "$30^\circ$" without a direction can lose the final accuracy mark.