This guide is for students learning kinematics and projectile motion in introductory physics, engineering undergraduates reviewing mechanics, and anyone needing worked solutions and sanity checks for lab reports.

“There are too many kinematic formulas; which one do I use?” “My answer is wrong when the projectile is launched from a cliff.” “How do I calculate landing speed without getting tangled in negative signs?”

Nearly all confusion surrounding projectile motion comes from mixing horizontal and vertical motions together. As Galileo first demonstrated, any projectile’s motion can be decomposed into two completely independent motions: horizontal motion at constant speed, and vertical motion accelerated downward by gravity.

This guide covers core concepts, how to choose the right equation for each point along the trajectory, quadratic equations for launches from heights, hand-calculated examples, and verification using a free browser tool.

Decision Table: Which Formula to Use and When

Equations in projectile motion follow directly from which specific point along the flight path you are investigating.

Target Point Physical Condition Target Quantity Formula to Use
Launch components Resolve vector by angle $\theta$ Horizontal initial $v_{0x}$
Vertical initial $v_{0y}$		 $v_{0x} = v_0 \cos\theta$ (constant)
$v_{0y} = v_0 \sin\theta$
Apex (Peak) Vertical velocity momentarily zero ($v_y = 0$) Time to peak $t_{\text{peak}}$
Maximum height $y_{\max}$		 $t_{\text{peak}} = \frac{v_{0y}}{g}$
$y_{\max} = h_0 + \frac{v_{0y}^2}{2g}$
Ground impact (Level ground) Height $y = 0$ with $h_0 = 0$ Flight duration $t_{\text{flight}}$
Horizontal range $R$		 $t_{\text{flight}} = \frac{2v_{0y}}{g} = 2t_{\text{peak}}$
$R = v_{0x} \times t_{\text{flight}} = \frac{v_0^2 \sin 2\theta}{g}$
Ground impact (Elevated cliff) Height $y = 0$ with $h_0 > 0$ Flight duration $t_{\text{flight}}$
Horizontal range $R$		 Positive root of $\frac{1}{2}gt^2 - v_{0y}t - h_0 = 0$
$R = v_{0x} \times t_{\text{flight}}$
Horizontal launch Launch angle $\theta = 0^\circ$ ($v_{0y} = 0$) Fall time $t$
Horizontal range $R$		 $t = \sqrt{\frac{2h_0}{g}}$
$R = v_0 \sqrt{\frac{2h_0}{g}}$
Impact velocity Vector addition or energy conservation Impact speed $v_{\text{land}}$ $v_{\text{land}} = \sqrt{v_x^2 + v_y^2} = \sqrt{v_0^2 + 2gh_0}$

Step 1: Resolve Initial Velocity into Components

Assuming negligible air resistance, the only force acting on the projectile after launch is gravity ($F_g = mg$), pointing straight down. There is zero horizontal force.

Consequently, the motion splits into two independent parts:

  1. Horizontal direction ($x$-axis): Zero net force $\to$ Uniform motion at constant speed ($v_x = v_{0x}$ throughout the entire flight).
  2. Vertical direction ($y$-axis): Constant downward acceleration $g$ $\to$ Uniformly accelerated motion (free fall / vertical throw).

Using trigonometry:

$$\begin{aligned} v_{0x} &= v_0 \cos\theta \ v_{0y} &= v_0 \sin\theta \end{aligned}$$

Unit Reminder: If initial speed is given in $\text{km/h}$, divide by $3.6$ to convert to $\text{m/s}$ before substituting into equations (e.g., $72\text{ km/h} \div 3.6 = 20\text{ m/s}$).

Step 2: Calculate the Apex (Peak Height)

At the highest point of the trajectory, the projectile temporarily stops rising. Its vertical velocity component momentarily equals zero ($v_y = 0$).

Setting $v_y = v_{0y} - gt = 0$:

$$t_{\text{peak}} = \frac{v_{0y}}{g} = \frac{v_0 \sin\theta}{g}$$

The peak height above ground level $y_{\max}$ combines initial elevation $h_0$ with vertical climb:

$$y_{\max} = h_0 + v_{0y} t_{\text{peak}} - \frac{1}{2}g (t_{\text{peak}})^2 = h_0 + \frac{v_{0y}^2}{2g}$$

Common Pitfall: “Velocity is zero at the peak” is Incorrect!

A frequent exam error is assuming the projectile’s overall speed is zero at the apex. Only the vertical component vanishes ($v_y = 0$). The horizontal component $v_x = v_0 \cos\theta$ remains entirely unaffected.

Thus, speed at the peak is non-zero: $v_{\text{peak}} = v_x = v_0 \cos\theta$ (only in a strictly vertical throw $\theta = 90^\circ$ does total velocity drop to zero).

Step 3: Flight Time and Horizontal Range

1. Level Ground Launch ($h_0 = 0$)

When launched and landed at the same elevation, the parabolic trajectory is perfectly symmetric.

  • Total Flight Time: Time ascending equals time descending: $$t_{\text{flight}} = 2 \times t_{\text{peak}} = \frac{2v_{0y}}{g} = \frac{2v_0 \sin\theta}{g}$$
  • Horizontal Range $R$: $$R = v_{0x} \times t_{\text{flight}} = (v_0 \cos\theta) \times \left(\frac{2v_0 \sin\theta}{g}\right) = \frac{v_0^2 (2\sin\theta \cos\theta)}{g} = \frac{v_0^2 \sin 2\theta}{g}$$

Why 45 Degrees Gives the Maximum Range

The term $\sin 2\theta$ reaches its maximum theoretical value of $1$ when $2\theta = 90^\circ$, which gives: $$\theta = 45^\circ$$ For launches over level ground with equal launch speeds, $45^\circ$ produces the furthest distance.

Furthermore, because $\sin(2 \times 30^\circ) = \sin 60^\circ$ and $\sin(2 \times 60^\circ) = \sin 120^\circ = \sin 60^\circ$, complementary launch angles that add up to $90^\circ$ (e.g., $30^\circ$ and $60^\circ$, or $20^\circ$ and $70^\circ$) achieve the exact same horizontal distance, though the steeper angle hangs in the air significantly longer.

2. Elevated Launch from a Cliff or Building ($h_0 > 0$)

When launching from an elevation, the landing surface is below the starting point ($y = 0$). The path is no longer symmetric.

Set vertical position $y(t) = h_0 + v_{0y}t - \frac{1}{2}gt^2 = 0$:

$$\frac{1}{2}gt^2 - v_{0y}t - h_0 = 0$$

Applying the quadratic formula for the positive root $t > 0$:

$$t_{\text{flight}} = \frac{v_{0y} + \sqrt{v_{0y}^2 + 2gh_0}}{g}$$

The horizontal distance follows:

$$R = v_{0x} \times t_{\text{flight}}$$

Why is the optimal launch angle from an elevation less than 45 degrees? The starting height $h_0$ naturally extends time in the air. Spending extra initial velocity pushing upward yields diminishing returns; investing more velocity into forward horizontal speed produces a longer total distance. Depending on cliff height, optimal launch angles typically range from $30^\circ$ to $40^\circ$.

Step 4: Horizontal Launch ($\theta = 0^\circ$)

A horizontal launch is a special case of projectile motion where $\theta = 0^\circ$:

  • Horizontal initial velocity: $v_{0x} = v_0 \cos 0^\circ = v_0$
  • Vertical initial velocity: $v_{0y} = v_0 \sin 0^\circ = 0$

Because vertical initial velocity is zero, the vertical motion is pure free fall from rest.

$$h_0 = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2h_0}{g}}$$

Horizontal range:

$$R = v_0 t = v_0 \sqrt{\frac{2h_0}{g}}$$

Galileo’s Thought Experiment: Fired Bullet vs. Dropped Bullet

If one bullet is fired horizontally from a cliff and another is dropped from rest from the same height at the exact same instant, which hits the ground first?

Ignoring air drag, they hit the ground at the exact same instant. Horizontal speed does not speed up or slow down vertical gravitational acceleration.

Step 5: Impact Velocity & Energy Conservation Check

You can compute landing speed $v_{\text{land}}$ via two independent methods:

Method A: Kinematic Velocity Components

  1. Horizontal velocity: $v_x = v_{0x}$ (constant)
  2. Vertical velocity at impact: $v_y = v_{0y} - g t_{\text{flight}}$ (negative, downward)
  3. Pythagorean combination: $v_{\text{land}} = \sqrt{v_x^2 + v_y^2}$

Method B: Conservation of Mechanical Energy (Fast Sanity Check)

Because gravity is a conservative force:

$$\frac{1}{2} m v_0^2 + mgh_0 = \frac{1}{2} m v_{\text{land}}^2 + 0$$

Canceling mass $m$ and solving for $v_{\text{land}}$:

$$v_{\text{land}} = \sqrt{v_0^2 + 2gh_0}$$

Notice that launch angle $\theta$ does not appear in this equation. Whether thrown upward, horizontally, or downward, any projectile launched at speed $v_0$ from height $h_0$ will strike the ground at the identical scalar speed. Use this equation to double-check your kinematic results.

Worked Examples

Example 1: 45° Launch on Level Ground

Problem: A ball is launched from ground level at $v_0 = 19.6\text{ m/s}$ at an angle of $45^\circ$. Taking $g = 9.8\text{ m/s}^2$, find:

  1. Time to peak and maximum height
  2. Total flight time and horizontal range
  3. Impact speed

Solution:

  1. Velocity Decomposition:
    • $v_{0x} = 19.6 \times \cos 45^\circ = 19.6 \times \frac{\sqrt{2}}{2} \approx 13.859\text{ m/s}$
    • $v_{0y} = 19.6 \times \sin 45^\circ = 19.6 \times \frac{\sqrt{2}}{2} \approx 13.859\text{ m/s}$
  2. Apex:
    • Time: $t_{\text{peak}} = \frac{v_{0y}}{g} = \frac{13.859}{9.8} = \sqrt{2} \approx \mathbf{1.414\text{ s}}$
    • Height: $y_{\max} = \frac{v_{0y}^2}{2g} = \frac{(13.859)^2}{19.6} = \mathbf{9.8\text{ m}}$
  3. Flight Time & Range:
    • Flight time: $t_{\text{flight}} = 2 \times \sqrt{2} \approx \mathbf{2.828\text{ s}}$
    • Range: $R = \frac{v_0^2 \sin 90^\circ}{g} = \frac{19.6^2}{9.8} = \mathbf{39.2\text{ m}}$
  4. Impact Speed:
    • Level ground means landing speed equals launch speed: $\mathbf{19.6\text{ m/s}}$ at $45^\circ$.

Example 2: Launch from an Elevated Cliff (Quadratic Equation)

Problem: A projectile is launched from a cliff of height $h_0 = 39.2\text{ m}$ with initial speed $v_0 = 14.0\text{ m/s}$ at an angle of $30^\circ$ above the horizontal ($g = 9.8\text{ m/s}^2$). Find:

  1. Maximum height above ground
  2. Total flight duration and horizontal range
  3. Impact speed

Solution:

  1. Components:
    • $v_{0x} = 14.0 \times \cos 30^\circ = 14.0 \times \frac{\sqrt{3}}{2} \approx 12.124\text{ m/s}$
    • $v_{0y} = 14.0 \times \sin 30^\circ = 14.0 \times 0.5 = 7.0\text{ m/s}$
  2. Peak Height:
    • Rise above cliff: $\Delta y = \frac{7.0^2}{2 \times 9.8} = \frac{49}{19.6} = 2.5\text{ m}$
    • Maximum height: $y_{\max} = 39.2 + 2.5 = \mathbf{41.7\text{ m}}$
  3. Flight Duration: Set vertical position to zero: $39.2 + 7.0t - 4.9t^2 = 0 \implies 4.9t^2 - 7.0t - 39.2 = 0$. Dividing by $4.9$: $t^2 - \frac{10}{7}t - 8 = 0$. $$t = \frac{\frac{10}{7} + \sqrt{\frac{100}{49} + 32}}{2} = \frac{10 + \sqrt{1668}}{14} \approx \frac{10 + 40.841}{14} \approx \mathbf{3.632\text{ s}}$$
  4. Horizontal Range: $$R = 12.124 \times 3.6315 \approx \mathbf{44.03\text{ m}}$$
  5. Impact Speed Verification (Energy Conservation): $$v_{\text{land}} = \sqrt{14.0^2 + 2(9.8)(39.2)} = \sqrt{196 + 768.32} = \sqrt{964.32} \approx \mathbf{31.05\text{ m/s}}$$

Verifying Worked Steps with the Online Calculator

To check your hand calculations or visualize trajectories for lab reports, use our free Projectile Motion Calculator.

Input Field Example Input Verified Output
Initial Speed $v_0$ 19.6 (supports m/s and km/h) Horizontal $v_{0x}$ and vertical $v_{0y}$ components
Launch Angle $\theta$ 45 (slider or numeric, $-89^\circ$ to $90^\circ$) Maximum height $y_{\max}$, time to peak $t_{\text{peak}}$
Initial Height $h_0$ 0 for ground, 39.2 for cliff Total flight time $t_{\text{flight}}$, horizontal range $R$
Gravity $g$ 9.8 (standard Earth), Moon, Mars, or custom Impact speed $v_{\text{land}}$, landing angle $\theta_{\text{land}}$
Time Slider $t$ Drag to inspect any flight instant Instantaneous $(x, y)$ coordinates and velocity vector

The tool generates an SVG trajectory graph, marks the apex and impact coordinates, and provides step-by-step mathematical working side-by-side with energy conservation verification.

Frequently Asked Questions

Q1. How does air resistance alter these results?

In real atmospheric conditions, aerodynamic drag acts proportional to velocity squared. This breaks the symmetry of the parabola: the projectile decelerates horizontally and descends steeper during the second half of its flight, forming a tear-drop curve. Both maximum height and horizontal range will be lower than the ideal theoretical values.

Q2. Why is object mass not included in the formulas?

In a vacuum, all objects fall with the exact same acceleration $g$, regardless of mass. In Newton’s second law $F = ma \implies mg = ma$, mass cancels out from both sides, leaving acceleration $a = g$.

Q3. How do I calculate downward throws (negative angles)?

Substitute the angle as a negative number (e.g., $\theta = -30^\circ$). The vertical initial velocity becomes downward ($v_{0y} = -v_0 \sin 30^\circ$). The launch point is the apex ($t_{\text{peak}} = 0$), and flight duration is obtained directly from the quadratic formula.

References & Standards