Physics · Kinematics

Projectile motion at an angle

Q: How do you find the maximum height in projectile motion?

For $vy_{0} > 0$, set $vy = vy_{0} - gt$ to zero. Then tpeak = vy₀/g and rise above launch is $H =\frac{ vy_{0}^{2}}{(2g)}$. With $vy_{0} =\frac{ 9 m}{s}$, H is 4.13 m. Absolute peak height is y₀ + H; equal launch and landing levels are not required. The horizontal velocity remains at the peak. For $vy_{0} \le{} 0$, the object does not rise after launch; its highest level in the considered flight is the initial one. A negative formal peak time lies before launch. No real intersection means the selected height is not reached in the model.

Q: What is the difference between angled and horizontal projectile motion?

A horizontal launch has $vy_{0} = 0$. From height h₀ above ground, $t = \sqrt{(\frac{2h_{0}}{g})}$ and $W = vx_{0}t$. With $h_{0} = 1.25 m$ and $vx_{0} =\frac{ 4 m}{s}$ this gives 0.505 s and 2.019 m. An angled upward launch additionally has a positive vertical initial component and can initially rise. Choose x rightward and y upward. v₀ is initial speed and α is measured from the positive horizontal. For the illustrated rightward motion, vx₀ = v₀cosα and vy₀ = v₀sinα. Initial position (x₀,y₀) must match the coordinate zero: $y_{0} = 0$ describes height relative to launch, not necessarily height above the ground.

At constant g, projectile motion separates free flight into constant horizontal velocity and vertical acceleration −g. The displayed coordinates use the launch point as origin.

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Formula

Calculate

x = v_{0} cos α \cdot t y = v_{0} sin α \cdot t - \frac{1}{2} g t^{2}

x, y in m · v₀ in m/s · α measured from horizontal (degrees or radians in the matching calculation mode) · t in s · local g in m/s²

Variables & units

Variables & units
Symbol	Meaning	Unit
$x$	Horizontal position at time t	m
$y$	Height above the launch point	m
$v₀$	Launch speed	m/s
$α$	Launch angle to the horizontal	°
$t$	Time since launch	s
$g$	Gravitational acceleration $(\frac{9.81 m}{s^{2}})$	m/s²

LaTeX

x = v_0 \cos\alpha \cdot t \qquad y = v_0 \sin\alpha \cdot t - \frac{1}{2} g t^2

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Diagram: a projectile parabola from the origin, the launch arrow v₀ at the angle α with its components v₀·cos α and v₀·sin α, at the apex an arrow g pointing down. — Ideal projectile parabola at constant g with no other forces. The illustrated special case has equal launch and landing heights; horizontal and vertical motion share the same time.

Projectile motion calculator

Find flight time, range, maximum height and impact speed of a throw to level ground, without air resistance.

Find

Launch speed v₀

Launch speed as a magnitude, e.g. 15 m/s for a strong ball throw.

Launch angle α

Launch angle above the horizontal, 0° to 90°, e.g. 30°; 0° is a horizontal throw.

Launch height above ground h

Height of the launch point above the ground, e.g. 2 m from your hand; 0 m when launched from the ground.

Gravitational acceleration g

Gravitational acceleration, 9.81 m/s² on Earth, about 1.62 m/s² on the Moon.

m/s²

Filled with example values. Type your own and the result updates straight away.

Result: Range R

R = 22.87 m

With v₀ = 15 m/s, α = 30 °, h = 2 m and g = 9.81 m/s², R = 22.87 m.

Working · rearrange for R, substitute, evaluate

1. Rearrange for R
$R=v_0\cos\alpha\,t$
2. Substitute values (in SI units)
- α = 30 ° = 0.523599 rad
$R=15\cdot \cos\left(0.523598775598\right)\cdot 1.76064=22.8714$
3. Result
R = 22.87 m

At h = 0 and vᵧ₀ > 0, the later return is selected; at h = vᵧ₀ = 0, contact is immediate. Impact speed does not determine contact force.

Quick tips

From the ground (h = 0) a throw at 45° goes furthest, and 30° and 60° land equally far. From a raised launch point the best angle is below 45°.
Split v₀ into v₀ · cos α horizontally and v₀ · sin α vertically: the horizontal speed stays the same, vertically only g acts.

Assumptions and limits

Launch from h ≥ 0 above ground, initial speed v₀ ≥ 0 and upward angle 0° to 90°; constant g > 0 with no air drag. Landing level is horizontal ground. Other landing heights and contact forces are outside this model.

Derivation & background

With x rightward and y upward, $v_x=v_0\cos\alpha$ and $v_y=v_0\sin\alpha-gt$ . Both position equations use the same time. The trajectory equation requires $vx_{0} \ne{} 0$ . Only with equal launch and landing levels and fixed v₀ is $W =\frac{ v_{0}^{2}\sin{}(2\alpha{})}{g}$ maximal at 45°. For an upward launch, rise $H =\frac{ v_{0}^{2}\sin{}^{2}\alpha{}}{(2g)}$ is independent of the later landing level.

Exam blueprint

Validity range

Valid with constant g and no drag or other forces. The compact position equation uses launch as the origin. $W =\frac{ v_{0}^{2}\sin{}(2\alpha{})}{g}$ and the 45° optimum require equal launch/landing levels and fixed v₀. For $vy_{0} > 0$ , $H =\frac{ vy_{0}^{2}}{(2g)}$ is rise above launch and does not require equal later landing height.

Derivation steps

Integrate constant gravity component by component and connect the motion through the same time.

1Choose x rightward and y upward; set initial position (x₀,y₀) and $t = 0$ .
2Resolve initial velocity: vx₀ = v₀cosα, vy₀ = v₀sinα.
3Use $ax = 0$ and $ay = -g$ in the ideal model; g is constant and positive.
4Integrate once: $vx = vx_{0}$ and $vy = vy_{0} - gt$ .
5Integrate again: $x = x_{0} + vx_{0}t$ and $y = y_{0} + vy_{0}t - \frac{1}{2}gt^{2}$ .
6For $vx_{0} \ne{} 0$ eliminate time: y − y₀ = (x − x₀)tanα $-\frac{ g(x - x_{0})^{2}}{(2vx_{0}^{2})}$ .
7Find landing time from $y = yL$ ; $W = vx_{0}t$ . Limit the 45° range rule to equal levels and fixed launch speed.

Rearrangements

Range

W = \frac{v_0^2 \sin(2\alpha)}{g}

Assumes equal launch and landing heights, fixed initial speed, an upward launch and negligible air forces. Only under these assumptions is range maximal at 45°.

Maximum height

H = \frac{v_0^2 \sin^2\alpha}{2g}

H is height gained above launch level for $v_{0}y > 0$ . Absolute peak height is y₀ + H; for $v_{0}y \le{} 0$ the highest point after launch is the starting position.

Time of flight

t = \frac{2 v_0 \sin\alpha}{g}

Holds for equal launch and landing height (twice the rise time).

Task variant

A ball launches at $v_{0} =\frac{ 15 m}{s}$ and 30° and lands at the same height. Neglect air forces and take $g =\frac{ 9.81 m}{s^{2}}$ . Find horizontal range.

W = v₀²·sin(2α)/g = 225 × sin(60°)/9.81 = 225 × 0.866/9.81 ≈ 19.9 m.

A ball launches at $v_{0} =\frac{ 20 m}{s}$ and 60°. Neglect air forces and use $g =\frac{ 9.81 m}{s^{2}}$ . Find the vertical height gained above launch level.

$H =\frac{ v_{0}^{2}\cdot{}\sin{}^{2}\alpha{}}{(2g)} =\frac{ 400 \times{} 0.75}{19.62} \approx{} 15.3 m$ , since sin²(60°) = 0.75.

Common mistakes

Using the full v₀ in one direction instead of splitting it into components.

Always decompose: v₀·cosα horizontally, v₀·sinα vertically.

Calculator set to radians while entering the angle in degrees.

Check the angle mode before calculating (DEG for degree values).

Applying the 45° rule for maximum range even with an elevated launch point.

A lower landing level can give an optimal angle below 45°; a higher target requires a different optimization with reachability constraints. Unequal heights alone do not establish a universal below-45° rule.

Assuming a horizontal acceleration.

Without air resistance the horizontal velocity stays constant.

Exam context

Typical tasks: computing range and maximum height, finding the flight time from the y equation or the landing point for an elevated launch, often combined with energy conservation.

Formula cluster

Projectile kinematics

Builds directly on free fall and uniformly accelerated motion.

h = ½gt²

Free Fall

s = ½at² + v₀t

Uniformly Accelerated Motion (Position–Time Equation)

v = a·t + v₀

Velocity-Time Law

Worked example

A ball launches at $v_0=20\,\mathrm{m/s}$ and $\alpha=45^\circ$ , landing at the same height. In the ideal model without air forces and with $g=9.81\,\mathrm{m/s^2}$ , use $W=\frac{v_0^2\sin(2\alpha)}g$ and $H=\frac{v_0^2\sin^2\alpha}{2g}$ . Substitution gives $W=\frac{400}{9.81}\,\mathrm m\approx40.8\,\mathrm m$ and height gained above launch $H=\frac{200}{19.62}\,\mathrm m\approx10.2\,\mathrm m$ .

Applications

Ball sports (launch angles), long jump and shot put, water fountains, ballistics, irrigation technology

How do you calculate with Projectile Motion (Trajectory Parabola)?

Here is how to work through a typical Projectile Motion (Trajectory Parabola) (x = v₀·cosα·t; y = v₀·sinα·t − ½gt²) task step by step:

1
Task
A ball launches at $v_{0} =\frac{ 15 m}{s}$ and 30° and lands at the same height. Neglect air forces and take $g =\frac{ 9.81 m}{s^{2}}$ . Find horizontal range.
Solution path
W = v₀²·sin(2α)/g = 225 × sin(60°)/9.81 = 225 × 0.866/9.81 ≈ 19.9 m.
2
Task
A ball launches at $v_{0} =\frac{ 20 m}{s}$ and 60°. Neglect air forces and use $g =\frac{ 9.81 m}{s^{2}}$ . Find the vertical height gained above launch level.
Solution path
$H =\frac{ v_{0}^{2}\cdot{}\sin{}^{2}\alpha{}}{(2g)} =\frac{ 400 \times{} 0.75}{19.62} \approx{} 15.3 m$ , since sin²(60°) = 0.75.

Understand and apply the equation

A thrown ball moves horizontally and vertically at the same time. In the ideal model horizontal velocity stays constant while gravity changes vertical velocity. Combining these motions determines flight time and trajectory, with unequal launch and landing heights treated explicitly.

Getting an intuitive picture

Imagine two views of the same flight. From above, the ball progresses steadily along the horizontal. From the side, you see its rise and subsequent fall. Both views describe the same object at the same time. You therefore cannot calculate horizontal distance using a separately chosen flight duration.[1]

Prerequisite: resolve a vector using sine and cosine. Here the angle is measured from the positive horizontal direction.
Use the free-fall equation for the vertical component and uniform motion for the horizontal. Both use the same elapsed time.
Specify launch point, landing level and axes. Assume constant g and no air forces. Contact is not part of the free-flight trajectory.

How do you resolve initial velocity and choose the axes?

Choose x rightward and y upward. v₀ is initial speed and α is measured from the positive horizontal. For the illustrated rightward motion, vx₀ = v₀cosα and vy₀ = v₀sinα. Initial position (x₀,y₀) must match the coordinate zero: y₀ = 0 describes height relative to launch, not necessarily height above the ground. Apply the velocity–time equation to each component with its own acceleration. For a horizontal launch, $v_{y0}=0$ ; vertical speed still grows afterwards under gravity.[1]

\begin{aligned}v_{0x}&=v_0\cos\alpha\\v_{0y}&=v_0\sin\alpha\\a_x&=0,\ a_y=-g\end{aligned}

g is a positive magnitude; the upward axis accounts for the minus sign.

Position is a point on the path; velocity is the instantaneous tangent direction. At the peak vy = 0 but vx remains. The object is therefore not stationary there when vx ≠ 0. Acceleration remains downward.

View diagram: force and motion components

One force, two motion components

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Weight points downward throughout ideal flight. At the peak, only vertical velocity vanishes, not force. Horizontal motion does not require continuous thrust in this model.[1]

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How do you derive the trajectory parabola from the equations of motion?

After launch, weight is the only force in this model. It acts vertically, so there is no horizontal force component and no horizontal acceleration. Initial velocity nevertheless has both components. Horizontal position follows from constant velocity times time; vertical position follows from initial velocity times time minus the falling contribution. The coordinate pair describes the ball’s position.[2]

\begin{aligned}x(t)&=x_0+v_{0x}t\\y(t)&=y_0+v_{0y}t-\frac12gt^2\end{aligned}

Evaluate both equations at the same time t. Initial height y₀ can be zero only when the chosen reference level is located there.[1]

\begin{aligned}v_x&=v_0\cos\alpha\\v_y&=v_0\sin\alpha-gt\end{aligned}

Differentiating position gives the velocity components.

y-y_0=(x-x_0)\tan\alpha-\frac{g(x-x_0)^2}{2v_0^2\cos^2\alpha}

For vx₀ ≠ 0 time can be eliminated; a purely vertical launch does not need this form.

Initial slope is tanα and the negative quadratic term curves the path downward. That term is in metres because g(x − x₀)²/v₀² is a length. Check angle mode: degree inputs require DEG, while programming libraries usually expect radians.

How do you find flight time, range and maximum height?

\begin{aligned}t_{\mathrm{peak}}&=\frac{v_{0y}}g\\H&=\frac{v_{0y}^2}{2g}\end{aligned}

For v₀y > 0, H is rise above the launch point, independent of the later landing level.

To find landing, first set y(t) = yLanding. The final level determines flight duration.
Choose the appropriate nonnegative time root. An obstacle ends flight at the first actual contact, not necessarily a later mathematical intersection.
Substitute that time into x(t). The horizontal component determines how far the ball travels during that same flight.

[1]

At equal launch and landing levels, y = y₀ gives the launch root t = 0 and return time 2v₀y/g. For a landing level Δy = yL − y₀ above or below launch, solve the full quadratic equation. Two positive roots may represent an upper level on ascent and descent; the actual obstacle determines which intersection matters. The vertical component follows free fall with the chosen initial velocity. Substitute the physically relevant landing time into horizontal motion only after selecting it.

\begin{aligned}t_{\pm}&=\frac{v_{0y}\pm\sqrt{v_{0y}^2-2g\Delta y}}g\\W&=v_{0x}t_{\mathrm{land}}\end{aligned}

Check the discriminant, t ≥ 0 and the first actual contact.

For vy₀ ≤ 0, the object does not rise after launch; its highest level in the considered flight is the initial one. A negative formal peak time lies before launch. No real intersection means the selected height is not reached in the model.

One clock for both components

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Trajectory from 2 m height at 15 m/s and 30°, without air resistance. Horizontal velocity stays constant while height first rises and then falls. The graph ends at first ground contact after about 1.76 s. Axes use different scales.[1]

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Example: components and an equal-level range

Return to the launch level

Given: vx₀ = 12 m/s, vy₀ = 9 m/s, x₀ = y₀ = 0; g = 9.81 m/s².
Find: Find launch speed, peak, flight time and range at the same landing level.

Approach: equal launch and landing heights allow return time 2v₀y/g. That time determines range through uniform horizontal motion; height gain depends only on the vertical component.
$\begin{aligned}t_{\mathrm{flight}}&=\frac{2v_{0y}}g\\R&=v_{0x}t_{\mathrm{flight}}\\H&=\frac{v_{0y}^2}{2g}\end{aligned}$
The position model explains why the same return time is used horizontally and vertically. At y = 0 the ball is back at launch level.
$\begin{aligned}x(t)&=v_{x0}t\\{}y(t)&=v_{y0}t-\frac12gt^2\\{}v_y(t)&=v_{y0}-gt\end{aligned}$
Combine components into magnitude and angle.
$\begin{aligned}v_0&=\sqrt{12^2+9^2}=15\,\mathrm{m/s}\\{} \alpha&=\arctan(\frac{9}{12})\approx36.87^\circ\end{aligned}$
Calculate the peak from vertical motion.
$\begin{aligned}t_{\mathrm{peak}}&=\frac{9}{9.81}\approx0.91743\,\mathrm s\\{} H&=\frac{81}{19.62}\approx4.12844\,\mathrm m\end{aligned}$
Find the later zero and horizontal displacement.
$\begin{aligned}t_{\mathrm{land}}&=\frac{18}{9.81}\approx1.83486\,\mathrm s\\{} W&=\frac{12\cdot18}{9.81}\approx22.01835\,\mathrm m\end{aligned}$

Result: Launch speed 15 m/s, angle 36.87°, peak height 4.13 m, flight time 1.83 s and range 22.02 m.

Check: At landing, components are (12,−9) m/s; speed is again 15 m/s. Energy is unchanged at the same level.

How do different launch and landing heights change flight time and impact speed?

The example launches two metres above landing level. Find landing time from the full vertical equation, then both velocity components. Kinetic energy together with the height change independently checks impact speed, but does not determine contact or peak force.[1][3]

A two-metre height difference

Given: v₀ = 15 m/s, α = 30°, y₀ = 2 m above level ground, x₀ = 0; g = 9.81 m/s².
Find: Calculate flight time, range and impact speed.

Approach: landing is below the start, so solve the general vertical equation. Then use the positive landing root for range and both velocity components.
$\begin{aligned}0&=y_0+v_{0y}t-\frac12gt^2\\t_{\mathrm{land}}&=\frac{v_{0y}+\sqrt{v_{0y}^2+2gy_0}}g\\R&=v_{0x}t_{\mathrm{land}}\end{aligned}$
Express the components using the specified angle. The upward-positive impact component is v₀sinα − gt and becomes negative at landing.
$\begin{aligned}0&=y_0+v_0\sin\alpha\,t-\frac12gt^2\\{}W&=v_0\cos\alpha\,t\\{}v_y&=v_0\sin\alpha-gt\end{aligned}$
Resolve launch velocity.
$\begin{aligned}v_{0x}&=15\cos30^\circ\approx12.99038\,\mathrm{m/s}\\{} v_{0y}&=7.5\,\mathrm{m/s}\end{aligned}$
Choose the positive root of y(t).
$t=\frac{7.5+\sqrt{7.5^2+2\cdot9.81\cdot2}}{9.81}\approx1.76064\,\mathrm s$
Use unrounded time for horizontal range.
$W\approx12.99038\cdot1.76064\approx22.87141\,\mathrm m$
Find the final vertical component and magnitude.
$\begin{aligned}v_{y,\mathrm{land}}&\approx-9.77190\,\mathrm{m/s}\\{} v_{\mathrm{land}}&=\sqrt{v_x^2+v_y^2}\approx16.25546\,\mathrm{m/s}\end{aligned}$

Result: Landing at 1.761 s and 22.871 m; impact speed 16.255 m/s.

Check: Energy gives vland² = v₀² + 2g·2 = 264.24 m²/s². The equal-level range formula would give only 19.86 m, incorrect for this landing level.

Explore further: when 45 degrees gives the greatest range

W=\frac{v_0^2\sin(2\alpha)}g\quad(y_L=y_0)

At fixed v₀ and equal levels, sin(2α) is maximal at α = 45°.

That optimum assumes equal launch and landing levels, negligible drag and launch speed independent of angle. For launch height h₀ above landing, the same ideal model gives tanαopt = v₀/√(v₀² + 2gh₀). With 15 m/s and 2 m in our example this is 42.70°. An elevated target or angle-dependent launch speed presents a different optimisation problem.

\tan\alpha_{\mathrm{opt}}=\frac{v_0}{\sqrt{v_0^2+2gh_0}}\quad(h_0\ge0)

This extension applies to a level landing surface below launch in the ideal model.

An original derivation uses y = h₀ + x tanα − gx²(1 + tan²α)/(2v₀²). Set u = tanα: at fixed x this equation is quadratic in u. At the greatest reachable x, the two angle solutions merge and the discriminant vanishes. This gives x² = v₀²(v₀² + 2gh₀)/g² and the stated optimum angle.

When is the ideal projectile parabola unsuitable?

Drag depends on velocity relative to air and usually has a horizontal component. The changes in vx and vy then become coupled, and the simple parabola is no longer exact. Wind, rotation and lift may also matter. Large altitude changes or long trajectories can also violate constant g.[4]

A full drawn parabola must not overlook an earlier obstacle. The first ground or wall contact ends free flight. Velocity immediately before it does not determine contact force; that needs an additional contact model.

Five solved exercises

Calculate · Foundation

vx₀ = 6 m/s, vy₀ = 8 m/s; equal landing level. Find launch speed and range.

Show hint

Find magnitude from components and time from vertical motion.

Learning goal: Find equal-height range.

Show solution and calculation

Approach: combine the components to obtain initial speed.
With equal launch and landing heights, flight time is twice the rise time.
The resulting range is about 9.79 m.

Check: on return vertical velocity is −8 m/s; combined with the unchanged horizontal component this gives speed 10 m/s again.

\begin{aligned}v_0&=\sqrt{v_{0x}^2+v_{0y}^2}=\sqrt{6^2+8^2}\,\mathrm{m/s}=10\,\mathrm{m/s}\\t&=\frac{2v_{0y}}g=\frac{16}{9.81}\,\mathrm s\\R&=v_{0x}t=\frac{6(16)}{9.81}\,\mathrm m\approx9.79\,\mathrm m\end{aligned}

[1]

Calculate · Application

A horizontal launch starts 1.25 m above ground with vx₀ = 4 m/s. Find flight time and range.

Show hint

vy₀ = 0; use the positive falling-time root.

Learning goal: Solve horizontal launch timing.

Show solution and calculation

Approach: a horizontal launch has zero initial vertical velocity, so falling time depends only on height.
For 1.25 m it is about 0.505 s.
Horizontal velocity stays at 4 m/s during that same time, giving range about 2.019 m.

Check: increasing horizontal velocity would change range but not falling time in this model.

\begin{aligned}t&=\sqrt{\frac{2h_0}g}=\sqrt{\frac{2(1.25)}{9.81}}\,\mathrm s\approx0.505\,\mathrm s\\R&=v_{0x}t\approx(4\,\mathrm{m/s})(0.50482\,\mathrm s)\\&\approx2.019\,\mathrm m\end{aligned}

[1]

Calculate · Application

For vx₀ = 10 m/s and vy₀ = 6 m/s, find velocity at 1 s.

Show hint

Keep the horizontal component and subtract gt vertically.

Learning goal: Find velocity vector after elapsed time.

Show solution and calculation

Approach: change only the vertical component by −gt, then calculate the velocity-vector magnitude.
At one second vertical velocity is −3.81 m/s and horizontal velocity remains 10 m/s.
Speed is about 10.70 m/s.

Check: the negative sign means downward motion; it does not make speed negative.

\begin{aligned}v_x&=10\,\mathrm{m/s}\\v_y&=6\,\mathrm{m/s}-(9.81\,\mathrm{m/s^2})(1\,\mathrm s)=-3.81\,\mathrm{m/s}\\|\vec v|&=\sqrt{v_x^2+v_y^2}=\sqrt{10^2+(-3.81)^2}\,\mathrm{m/s}\\&\approx10.70\,\mathrm{m/s}\end{aligned}

[1]

Interpret · Application

An ideal projectile without drag starts with vx0=5 m/s and vy0=7 m/s; g=9.81 m/s² downward. At its highest point an observer calls it completely at rest. State its velocity and acceleration there.

Show hint

Only the vertical velocity component vanishes.

Learning goal: Interpret a trajectory peak using components.

Show solution and calculation

Approach: The peak is defined by zero vertical velocity while horizontal velocity remains unchanged.
The projectile still moves horizontally at 5 m/s; acceleration stays 9.81 m/s² downward.
Peak time is approximately 0.71356 s.
A horizontal tangent does not mean zero net force.

\begin{aligned}t_{\mathrm{peak}}&=\frac{v_{y0}}g\approx0.71356\,\mathrm s\\{}\vec v_{\mathrm{peak}}&=(5,0)\,\mathrm{m\,s^{-1}}\\{}\vec a&=(0,-9.81)\,\mathrm{m\,s^{-2}}\end{aligned}

[1]

Error analysis · Application

A body is launched horizontally at 6 m/s from height 2 m; g=9.81 m/s², no drag. Someone applies the equal-height range formula and gets zero because sin(0)=0. Diagnose the model error and find actual ground range.

Show hint

Ground is below launch level; solve vertical fall first.

Learning goal: Check the equal-height assumption of the projectile range formula.

Show solution and calculation

Approach: The equal-height formula requires returning to launch level.
Here obtain positive fall time from the vertical position equation, then multiply by constant horizontal velocity.
Flight time is about 0.63855 s and range about 3.83131 m.
Ground lies below launch, so zero range does not solve this boundary-value problem.

\begin{aligned}0&=2\,\mathrm m-\frac12gt^2\\{}t&=\sqrt{\frac{4\,\mathrm m}{g}}\approx0.63855\,\mathrm s\\{}W&=v_{x0}t\approx3.83131\,\mathrm m\end{aligned}

[1]

What to take away

One flight duration links horizontal and vertical motion. First identify the final level, then calculate time and finally range.
At the peak only vᵧ vanishes. Horizontal velocity and downward acceleration remain.
The 45-degree rule requires equal heights, fixed initial speed and negligible air forces. A higher or lower target changes the optimization problem.

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 2 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Made components, trajectory, relevant landing time and unequal heights easier to find; clarified horizontal launch and linked motion, free fall and the energy check in context.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added prerequisites and an intuitive introduction; clarified the derivation, worked-example approaches and exercise solutions, and added a summary.

Cite this reference page

Quanta. Projectile Motion (Trajectory Parabola). Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/schiefer-wurf

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Frequently asked questions about Projectile Motion (Trajectory Parabola)

How do you calculate the range of projectile motion?+

Find landing time from the vertical equation first, then calculate $W = vx_{0}t$ . With equal levels and upward initial motion, $t =\frac{ 2vy_{0}}{g}$ . For $vx_{0} =\frac{ 12 m}{s}$ and $vy_{0} =\frac{ 9 m}{s}$ this gives $t = 1.835 s$ and $W = 22.02 m$ . The familiar range formula is a special case of this method. At equal launch and landing levels, $y = y_{0}$ gives the launch root $t = 0$ and return time $\frac{2v_{0}y}{g}$ . For a landing level $\Delta{}y = yL - y_{0}$ above or below launch, solve the full quadratic equation. Two positive roots may represent an upper level on ascent and descent; the actual obstacle determines which intersection matters.

Why is 45 degrees the optimal launch angle?+

With equal levels, W contains sin(2α), which reaches one at $\alpha{} = 45$ ° for fixed v₀. Shallower angles increase the horizontal launch component while steeper angles increase flight time; the exact 45° optimum belongs to these assumptions. It is not a universal recommendation for sporting throws or real projectiles. That optimum assumes equal launch and landing levels, negligible drag and launch speed independent of angle. For launch height h₀ above landing, the same ideal model gives tanαopt = v₀/√(v₀² + 2gh₀). With $\frac{15 m}{s}$ and 2 m in our example this is 42.70°. An elevated target or angle-dependent launch speed presents a different optimisation problem.

How do you find the maximum height in projectile motion?+

For $vy_{0} > 0$ , set $vy = vy_{0} - gt$ to zero. Then tpeak = vy₀/g and rise above launch is $H =\frac{ vy_{0}^{2}}{(2g)}$ . With $vy_{0} =\frac{ 9 m}{s}$ , H is 4.13 m. Absolute peak height is y₀ + H; equal launch and landing levels are not required. The horizontal velocity remains at the peak. For $vy_{0} \le{} 0$ , the object does not rise after launch; its highest level in the considered flight is the initial one. A negative formal peak time lies before launch. No real intersection means the selected height is not reached in the model.

What is the difference between angled and horizontal projectile motion?+

A horizontal launch has $vy_{0} = 0$ . From height h₀ above ground, $t = \sqrt{(\frac{2h_{0}}{g})}$ and $W = vx_{0}t$ . With $h_{0} = 1.25 m$ and $vx_{0} =\frac{ 4 m}{s}$ this gives 0.505 s and 2.019 m. An angled upward launch additionally has a positive vertical initial component and can initially rise. Choose x rightward and y upward. v₀ is initial speed and α is measured from the positive horizontal. For the illustrated rightward motion, vx₀ = v₀cosα and vy₀ = v₀sinα. Initial position (x₀,y₀) must match the coordinate zero: $y_{0} = 0$ describes height relative to launch, not necessarily height above the ground.

Why may the two motions be treated separately?+

Gravity has no horizontal component in the chosen coordinates, giving $ax = 0$ and $ay = -g$ . Both motion equations use compatible initial data and the same time; together their components determine one velocity vector and one trajectory. The separation does not introduce two different clocks or two different objects. Drag depends on velocity relative to air and usually has a horizontal component. The changes in vx and vy then become coupled, and the simple parabola is no longer exact. Wind, rotation and lift may also matter. Large altitude changes or long trajectories can also violate constant g.

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[1]OpenStax, University Physics Vol. 1, §4.3 Projectile Motion

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Projectile motion at an angle

Variables & units

Projectile motion calculator

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Projectile kinematics

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How do you calculate with Projectile Motion (Trajectory Parabola)?

Getting an intuitive picture

How do you resolve initial velocity and choose the axes?

One force, two motion components

How do you derive the trajectory parabola from the equations of motion?

How do you find flight time, range and maximum height?

One clock for both components

Example: components and an equal-level range

Return to the launch level

How do different launch and landing heights change flight time and impact speed?

A two-metre height difference

Explore further: when 45 degrees gives the greatest range

When is the ideal projectile parabola unsuitable?

Five solved exercises

vx0 = 6 m/s, vy0 = 8 m/s; equal landing level. Find launch speed and range.

A horizontal launch starts 1.25 m above ground with vx0 = 4 m/s. Find flight time and range.

For vx0 = 10 m/s and vy0 = 6 m/s, find velocity at 1 s.

An ideal projectile without drag starts with vx0=5 m/s and vy0=7 m/s; g=9.81 m/s2 downward. At its highest point an observer calls it completely at rest. State its velocity and acceleration there.

A body is launched horizontally at 6 m/s from height 2 m; g=9.81 m/s2, no drag. Someone applies the equal-height range formula and gets zero because sin(0)=0. Diagnose the model error and find actual ground range.

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Retain Projectile Motion (Trajectory Parabola) for exams

vx₀ = 6 m/s, vy₀ = 8 m/s; equal landing level. Find launch speed and range.

A horizontal launch starts 1.25 m above ground with vx₀ = 4 m/s. Find flight time and range.

For vx₀ = 10 m/s and vy₀ = 6 m/s, find velocity at 1 s.

An ideal projectile without drag starts with vx0=5 m/s and vy0=7 m/s; g=9.81 m/s² downward. At its highest point an observer calls it completely at rest. State its velocity and acceleration there.

A body is launched horizontally at 6 m/s from height 2 m; g=9.81 m/s², no drag. Someone applies the equal-height range formula and gets zero because sin(0)=0. Diagnose the model error and find actual ground range.