Maximum Height Calculator – Projectile Motion

Maximum height in projectile motion

Use this maximum height calculator to figure out what is the maximum vertical position of an object in projectile motion.

Last updated: December 29, 2025
Frank Zhao - Creator
CreatorFrank Zhao
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Projectile motion: maximum height
1Horizontal velocity component
vx=vcos(α)v_x = v \cos(\alpha)
2Vertical velocity component
vy=vsin(α)v_y = v \sin(\alpha)
3Maximum height
hmax=h0+vy22g=h0+v2sin2(α)2gh_{\max} = h_0 + \frac{v_y^2}{2g} = h_0 + \frac{v^2 \sin^2(\alpha)}{2g}
vvVelocity
vxv_xHorizontal velocity
vyv_yVertical velocity
α\alphaAngle of launch
h0h_0Initial height
hmaxh_{\max}Maximum height

Introduction / Overview

The Maximum Height Calculator is a tool for finding the highest vertical position a launched object reaches during projectile motion. Whether you throw a ball, kick a soccer shot, or run a classroom experiment, it helps you answer one simple question: “How high will it go?”

What problems does it solve?

It quickly links your inputs (launch speed, angle, and initial height) to peak height. It can also work in reverse: if you know the peak height, it can help you back-calculate other values.

Who is this for?

Students checking homework, athletes and coaches sanity-checking a throw, makers building launchers, and anyone who wants a fast, unit-safe answer.

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Reliability note:

The calculator uses standard constant-acceleration physics with gravity. It assumes no air resistance and a constant gravitational acceleration.

How to find the maximum height of a projectile?

A projectile rises until gravity slows the vertical motion to a stop. At the very top of the trajectory, the vertical velocity is:vy=0v_y = 0.

The 3 key ideas (no heavy math):

  • Split the launch speed into components:vy0=v0sin(α)v_{y0} = v_0\sin(\alpha).
  • Time to the peak happens whenvy(t)=vy0gt=0v_y(t) = v_{y0} - g t = 0.
  • Height gained depends on the vertical part only (the horizontal speed doesn’t lift the projectile).

A quick mental shortcut

If you launch atα=0\alpha = 0^\circ(perfectly horizontal), thenvy0=0v_{y0} = 0and the projectile never rises above the launch height. So:hmax=h0h_{\max} = h_0.

Maximum height calculator helps you find the answer

You can use this calculator in a “textbook” direction (give it the launch conditions and read the peak height), or in a “reverse” direction (give it peak height and solve for missing inputs). Either way, it’s designed for quick what-if experiments.

1

Enter the launch velocity

Type the initial speed and choose units (for example, ft/s or m/s).

2

Enter the launch angle

Use degrees or radians. The calculator handles conversions. Remember:α=90\alpha = 90^\circis straight up.

3

(Optional) Enter initial height

If you launch from a platform, enter the starting height. If you launch from the ground, set it to 0.

4

Read the maximum height

The maximum height updates instantly. If you change any input, the peak height will follow.

Screenshot / diagram

Projectile motion diagram for maximum height

The diagram is a visual reminder: only the vertical motion determines the peak height.

Example 1: Can a 30 ft/s kick clear a 13 ft fence?

Suppose you kick a ball atv0=30 ft/sv_0 = 30\ \mathrm{ft/s}with an angle ofα=70\alpha = 70^\circfrom ground levelh0=0 fth_0 = 0\ \mathrm{ft}. Ignoring air resistance, the peak height is:

hmax=h0+v02sin2(α)2gh_{\max} = h_0 + \frac{v_0^2\sin^2(\alpha)}{2g}==0+302sin2(70)232.1740 + \frac{30^2\sin^2(70^\circ)}{2\cdot 32.174}\approx12.35 ft12.35\ \mathrm{ft}

So it tops out at about 12.35 ft — just under the 13 ft fence. To clear it, you’d need a bit more speed, or a slightly different angle.

Example 2: Launching from a 2 m platform

Now let’s use SI units:v0=20 m/sv_0 = 20\ \mathrm{m/s},α=45\alpha = 45^\circ, andh0=2 mh_0 = 2\ \mathrm{m}.

hmaxh_{\max}==2+202sin2(45)29.806652 + \frac{20^2\sin^2(45^\circ)}{2\cdot 9.80665}\approx12.20 m12.20\ \mathrm{m}

The projectile peaks at about 12.2 m above the ground. Notice how the initial height simply shifts the final answer upward.

Real-World Examples / Use Cases

Sports (kicks and throws)

Inputs: v0=25 m/s, α=55, h0=0v_0 = 25\ \mathrm{m/s},\ \alpha = 55^\circ,\ h_0 = 0

Result: hmax26.0 mh_{\max} \approx 26.0\ \mathrm{m}

Great for quick “will it clear the bar?” checks.

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Launching from a platform

Inputs: v0=18 m/s, α=40, h0=8 mv_0 = 18\ \mathrm{m/s},\ \alpha = 40^\circ,\ h_0 = 8\ \mathrm{m}

Result: hmax=h0+v02sin2α2gh_{\max} = h_0 + \frac{v_0^2\sin^2\alpha}{2g}

Useful in labs and simple engineering demos.

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Game development tuning

Inputs: v0 and αv_0 \text{ and } \alpha

Result: hmaxh_{\max}

Balance “arc height” so jumps feel right.

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Homework sanity checks

Inputs: v0, α, h0v_0,\ \alpha,\ h_0

Result: hmax and thh_{\max} \text{ and } t_h

Cross-check hand calculations quickly.

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Classroom experiments

Inputs: hmax observedh_{\max} \text{ observed}

Result: vy0=2g(hmaxh0)v_{y0} = \sqrt{2g(h_{\max}-h_0)}

Back-calculate vertical launch speed from data.

Want the full picture (range + time + max height)? Try the Projectile Motion Calculator.

Common Scenarios / When to Use

  • You know speed and angle

    This is the classic setup. Enterv0v_0andα\alphaand readhmaxh_{\max}.

  • You measured peak height

    If you can estimatehmaxh_{\max}(video analysis, a marked wall, or a sensor), the calculator can help infer vertical launch speed.

  • You’re studying special angles

    Forα=90\alpha = 90^\circthe motion is purely vertical. Forα=0\alpha = 0^\circit becomes a horizontal launch.

  • You’re in strong air resistance

    Golf balls, feathers, long-range ballistics, and high-speed projectiles can deviate a lot from the no-drag model. In that case, treat this calculator as a baseline approximation.

Tips & Best Practices

Use realistic angles

Most real throws are somewhere between2020^\circand7070^\circ. If you type extreme angles, the calculator is still correct, but the scenario may not match reality.

Sanity-check units

Peak height scales withv02v_0^2. If your answer looks 10× too big, it’s often a unit mix-up (mph vs m/s, ft vs m).

Remember: only vertical speed lifts

If you keepvy0v_{y0}the same (even if you change the horizontal speed), the peak height stays the same.

Pair with range/time tools

If your next question is “How far?” or “How long?”, jump toRangeorTime of flightcalculators.

Important: this calculator ignores air resistance. If your object is very light or slow (like a shuttlecock), real-life peak height will likely be lower than the idealized estimate.

Calculation Method / Formula Explanation

The maximum height comes from the vertical motion with constant accelerationg-g. Start with the vertical velocity equation:

vy(t)=vy0gtv_y(t) = v_{y0} - g t

Variables (what each symbol means)

  • v0v_0: initial speed
  • α\alpha: launch angle
  • h0h_0: initial height
  • hmaxh_{\max}: maximum height
  • gg: gravitational acceleration

At the highest point,vy=0v_y = 0. Solve for the time to peak:

0=vy0gth0 = v_{y0} - g t_h\Rightarrowth=vy0g=v0sin(α)gt_h = \frac{v_{y0}}{g} = \frac{v_0\sin(\alpha)}{g}

The vertical displacement equation is:

y(t)=vy0tgt22y(t) = v_{y0} t - \frac{g t^2}{2}

Plug int=tht = t_hand add the initial height:

hmax=h0+vy0thgth22h_{\max} = h_0 + v_{y0} t_h - \frac{g t_h^2}{2}==h0+vy022gh_0 + \frac{v_{y0}^2}{2g}==h0+v02sin2(α)2gh_0 + \frac{v_0^2\sin^2(\alpha)}{2g}

Special cases (quick intuition)

  • If α=90\alpha = 90^\circ, then sin(α)=1\sin(\alpha)=1 and the formula becomes hmax=h0+v022gh_{\max} = h_0 + \frac{v_0^2}{2g}.
  • If α=45\alpha = 45^\circ, then sin2(α)=12\sin^2(\alpha)=\frac{1}{2} and hmax=h0+v024gh_{\max} = h_0 + \frac{v_0^2}{4g}.
  • If α=0\alpha = 0^\circ, then hmax=h0h_{\max}=h_0 (no upward component).

Related Concepts / Background Info

Why max height depends on the vertical part

Projectile motion is two independent motions: a horizontal motion with (approximately) constant speed, and a vertical motion with constant acceleration. The peak height is controlled byvy0v_{y0}andggonly.

Mass doesn’t matter (in this model)

With no drag, the equations for height don’t include mass. In real life, mass can matter indirectly because it changes how air resistance affects the object.

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Next step ideas:

Once you havehmaxh_{\max}, you may want to compute the full flight path or landing distance. Use the Projectile Range Calculator or the Projectile Motion Calculator.

FAQs

How do I find the maximum height of a ball thrown up?

Usehmax=h0+vy022gh_{\max} = h_0 + \frac{v_{y0}^2}{2g}wherevy0=v0sin(α)v_{y0}=v_0\sin(\alpha). If you throw straight up, thenα=90\alpha=90^\circandvy0=v0v_{y0}=v_0.

What angle has the longest flight time (ignoring air resistance)?

For a fixed launch speed, flight time depends on the vertical component, so it’s longest atα=90\alpha = 90^\circ. A common approximation ist2v0sin(α)gt \approx \frac{2v_0\sin(\alpha)}{g}when launching and landing at the same height.

Does mass affect maximum height?

In the ideal no-drag model: no. The peak height depends onv0v_0,α\alpha,h0h_0, andgg. With air resistance, different masses and shapes can behave differently.

What factors affect projectile motion in real life?

The big one is aerodynamic drag (air resistance). Wind, spin (Magnus effect), and even changingggslightly with altitude can matter in precise applications.

Why does a higher angle usually increase max height?

Because the vertical componentvy0=v0sin(α)v_{y0}=v_0\sin(\alpha)increases withα\alpha(up to9090^\circ), andhmaxh0vy02h_{\max}-h_0 \propto v_{y0}^2.

What if I only know the max height — can I solve for launch speed?

Yes. Rearrange the formula:vy0=2g(hmaxh0)v_{y0}=\sqrt{2g(h_{\max}-h_0)}. Then usev0=vy0sin(α)v_0 = \frac{v_{y0}}{\sin(\alpha)}if you also knowα\alpha.

Why is my measured maximum height lower than the calculator?

Most commonly: air resistance, launch speed overestimated, or the launch angle is not what you think. Small angle errors matter becausesin2(α)\sin^2(\alpha)changes quickly at steep angles.

Is this calculator accurate for very long distances or bullets?

Use it as a baseline only. Long-range ballistics require drag models, and sometimes Earth curvature and varying air density.

Limitations / Disclaimers

This calculator uses an ideal projectile model: no air resistance, no wind, and constant gravity. Use it for education, planning, and rough estimates — not as a substitute for professional engineering or safety analysis.

External References / Sources

  • Calculus: Early Transcendentals (Ninth Edition)

    Stewart, James; Clegg, Dan; Watson, Saleem (2021)
    ISBN: 978-1-337-61392-7 | p.919

  • Doing Physics with Scientific Notebook: A Problem Solving Approach

    Gallant, Joseph (2012)
    ISBN: 978-1-119-94194-1 | p.132
    View on Google Books →

  • Classical Mechanics

    Tatum (2019) | Chapter 7
    PDF Link →

  • Classical Dynamics of Particles and Systems

    Stephen T. Thornton; Jerry B. Marion (2007)
    ISBN: 978-0-495-55610-7 | p.59
    View on Google Books →

  • Classical Mechanics: Point Particles and Relativity

    Walter Greiner (2004)
    ISBN: 0-387-95586-0 | p.181
    View on Google Books →

  • Two New Sciences

    Galileo Galilei (1638) | p.249
    Historic original research on projectile motion

Maximum Height Calculator – Projectile Motion