Projectile Motion Equations: Worked Examples from Ground and Cliff

A ball leaves the ground at 20 m/s, 30 degrees above the horizontal, and lands about 35 m away. Throw the same ball the same way from the top of a 15 m cliff and it lands about 53 m out. Plenty of students get 35 m for the cliff too, because they reached for the range formula without checking what it assumes. The projectile motion equations are not hard. Knowing which ones apply to the launch in front of you is the whole skill.

This guide sets up one sign convention, puts every equation in one table with the condition it needs, and then works the same launch from level ground and from a cliff, plus a horizontal launch from a table. All numbers use $g=9.81\ \mathrm{m/s^2}$ and no air resistance.

Split the launch into two motions

Sketch first. Draw the launch point, the ground, and an arrow for the initial velocity $v_0$ at angle $\theta$ above the horizontal. Put the origin at the launch point, take up as positive and forward as positive. Then resolve the launch velocity:

$$v_{0x}=v_0\cos\theta,\qquad v_{0y}=v_0\sin\theta.$$

From here on, the problem is two one-dimensional problems that share a clock. Horizontally, nothing pushes the ball once it is released, so $a_x=0$ and $v_x$ stays at $v_{0x}$. Vertically, gravity gives a constant $a_y=-g$. OpenStax's projectile motion section builds the topic on exactly this independence of perpendicular motions, with air resistance neglected. The AP Physics 1 course framework says the same thing in one line: projectile motion is two-dimensional motion with zero acceleration in one direction and constant, nonzero acceleration in the other (College Board, AP Physics 1 CED, topic 1.5).

If the split itself still feels shaky, our kinematics guide covers the one-dimensional constant-acceleration work that each axis uses.

Which equation is valid here?

Four equations describe any projectile in this model. Three more are shortcuts derived from them, and each shortcut carries a condition that is easy to forget.

Equation	What it gives	Valid when
$x=v_{0x}t$	horizontal position	always (no air resistance)
$y=y_0+v_{0y}t-\tfrac{1}{2}gt^2$	height at time $t$	always
$v_x=v_{0x}$	horizontal velocity	always
$v_y=v_{0y}-gt$	vertical velocity	always
$T=\dfrac{2v_0\sin\theta}{g}$	time of flight	lands at the same height it left
$H=\dfrac{v_0^2\sin^2\theta}{2g}$	maximum height	measured above the launch point; holds at any landing height
$R=\dfrac{v_0^2\sin 2\theta}{g}$	range	lands at the same height it left

OpenStax states both limits directly: the time-of-flight result does not apply when the projectile lands at a different elevation, and the range equation is valid only for launch and impact on a horizontal surface. The maximum height formula survives a cliff because it only uses the vertical motion up to the top of the arc, where $v_y=0$. It tells you how high the ball rises above where it started, not above the ground below.

Eliminating $t$ from the first two rows gives the trajectory, a parabola:

$$y=x\tan\theta-\frac{g\,x^2}{2v_0^2\cos^2\theta}.$$

It is handy as an independent check, as you will see in the cliff example.

Example 1: level ground at 20 m/s and 30°

Components first: $v_{0x}=20\cos 30^\circ=17.32\ \mathrm{m/s}$ and $v_{0y}=20\sin 30^\circ=10.0\ \mathrm{m/s}$.

Time of flight. Launch and landing are at the same height, so the shortcut is allowed:

$$T=\frac{2(10.0)}{9.81}=2.04\ \mathrm{s}.$$

Check another way: the ball reaches the top when $v_y=0$, at $t=10.0/9.81=1.02\ \mathrm{s}$. The way down mirrors the way up on level ground, so the total is twice that, 2.04 s.

Maximum height.

$$H=\frac{(10.0)^2}{2(9.81)}=5.10\ \mathrm{m}.$$

Check by substituting the peak time into the height equation: $y=10.0(1.02)-4.905(1.02)^2=5.10\ \mathrm{m}$.

Range.

$$R=\frac{(20)^2\sin 60^\circ}{9.81}=35.3\ \mathrm{m}.$$

Check without the shortcut: $x=v_{0x}T=17.32\times 2.04=35.3\ \mathrm{m}$. Two methods, same answer, which is what you want before moving on. If your course uses $g=9.8\ \mathrm{m/s^2}$ (AP Physics 1 practice in Physics Zen does), the answers shift only in the third figure: 2.04 s, 5.10 m and 35.3 m all round the same.

Example 2: the same launch from a 15 m cliff

Now launch at the same 20 m/s and 30°, but from the edge of a cliff 15 m above flat ground. With the origin at the launch point, landing happens at $y=-15\ \mathrm{m}$. The range shortcut is off the table, because the ball lands lower than it started. Use the general height equation:

$$-15=10.0\,t-4.905\,t^2.$$

Rearranged, $4.905t^2-10.0t-15=0$. The quadratic formula gives

$$t=\frac{10.0\pm\sqrt{(10.0)^2+4(4.905)(15)}}{2(4.905)}=\frac{10.0\pm 19.86}{9.81},$$

so $t=3.04\ \mathrm{s}$ or $t=-1.00\ \mathrm{s}$. The negative root is a moment before the throw, when a ball on the same parabola would have been passing through ground level on its way up. It is valid algebra and irrelevant physics, so reject it. Substitute the kept root back to confirm: $10.0(3.044)-4.905(3.044)^2=-15.0$. Root choice is one of the math slips we cover in physics mistakes that are really math mistakes.

Range from the cliff.

$$x=17.32\times 3.044=52.7\ \mathrm{m}.$$

Second check with the trajectory equation: at $x=52.7\ \mathrm{m}$, $y=52.7\tan 30^\circ-\dfrac{9.81(52.7)^2}{2(20)^2\cos^2 30^\circ}=30.4-45.4=-15.0\ \mathrm{m}$. The ball is exactly at the ground.

What the shortcut would have cost. Reusing $R=v_0^2\sin 2\theta/g$ here gives 35.3 m, the level-ground answer. The true range is 52.7 m. The shortcut is 17.4 m short, about 33 percent of the real distance, and nothing in the formula warns you. The maximum height is still 5.10 m above the launch point, which is 20.1 m above the ground below.

Impact velocity. At landing, $v_y=10.0-9.81(3.044)=-19.86\ \mathrm{m/s}$ and $v_x$ is still 17.32 m/s. So

$$v=\sqrt{17.32^2+19.86^2}=26.3\ \mathrm{m/s},$$

directed $\arctan(19.86/17.32)=48.9^\circ$ below the horizontal. Energy gives the speed independently: $v=\sqrt{v_0^2+2gh}=\sqrt{20^2+2(9.81)(15)}=26.3\ \mathrm{m/s}$. Our work and energy guide explains why that shortcut works.

Is 45° still the best angle? Not from a cliff. Keeping 20 m/s and the 15 m drop, we solved the same quadratic at several angles:

Launch angle	30°	35°	37°	40°	45°
Range from the 15 m cliff	52.7 m	53.6 m	53.7 m	53.6 m	52.4 m

The best of these is 37°, and 45° lands shorter than 30°. A raised launch already buys extra time in the air, so a flatter, faster horizontal launch pays more.

Example 3: a ball rolling off a table

A ball rolls off a table 0.90 m high at 3.0 m/s. Here $\theta=0$, so $v_{0x}=3.0\ \mathrm{m/s}$ and $v_{0y}=0$. Again the landing height differs from the launch height, so only the general equations apply.

Time to fall. With $v_{0y}=0$, the height equation becomes $-0.90=-4.905t^2$:

$$t=\sqrt{\frac{2(0.90)}{9.81}}=0.428\ \mathrm{s}.$$

This is the same fall time as a ball dropped straight down from the table edge. The horizontal speed does not slow the fall.

Where it lands. $x=3.0\times 0.428=1.29\ \mathrm{m}$ from the foot of the table. LibreTexts' projectile motion page treats the horizontal launch from height the same way, with fall time $\sqrt{2H/g}$.

Speed and angle at impact. $v_y=-9.81(0.428)=-4.20\ \mathrm{m/s}$, so

$$v=\sqrt{3.0^2+4.20^2}=5.16\ \mathrm{m/s},\qquad \alpha=\arctan\frac{4.20}{3.0}=54.5^\circ$$

below the horizontal. Energy agrees: $\sqrt{3.0^2+2(9.81)(0.90)}=5.16\ \mathrm{m/s}$.

Two practice problems, with answers

1. Level ground. A ball is kicked at 15 m/s, 60° above level ground. Find the flight time, maximum height and range. Answers: $v_{0y}=13.0\ \mathrm{m/s}$, $T=2.65\ \mathrm{s}$, $H=8.60\ \mathrm{m}$, $R=19.9\ \mathrm{m}$. All three shortcuts are allowed here; check the range with $v_{0x}T=7.5\times 2.65$.

2. Horizontal launch. A stone is thrown horizontally at 12 m/s from a ledge 20 m above the water. How long is it in the air, how far out does it land, and how fast is it going on impact? Answers: $t=\sqrt{2(20)/9.81}=2.02\ \mathrm{s}$, $x=24.2\ \mathrm{m}$, $v=\sqrt{12^2+2(9.81)(20)}=23.2\ \mathrm{m/s}$ at $58.8^\circ$ below the horizontal. No shortcut from the level-ground rows applies.

Practice the split until it is automatic

Projectiles appear early in every major course: AP Physics 1 places them in Unit 1, and our AP Physics 1 guide shows where kinematics sits in a full study plan. The method above does not change between exams; only $g$ and the numbers do.

The Kinematics topic in Physics Zen, free end to end on iPhone and Android, includes a projectile motion skill with fresh launch speeds, angles and heights each time, instant checking, and a worked solution after every answer. Do a handful, and before each one ask the question from the table: does this ball land at the height it left? If not, skip the shortcuts and solve for time.