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Physics

Projectile Motion

A thrown ball moves forward while gravity pulls it downward. Explore how these two motions combine into a parabolic path, then predict its height, flight time, and landing point.
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One flight, two independent motions

Projectile motion is free flight under gravity after an object is launched. Take rightward as positive xx and upward as positive yy. With air resistance neglected, gravity affects the vertical motion while the horizontal velocity remains unchanged.

ax=0,ay=−ga_x=0,\qquad a_y=-g

The same elapsed time connects both directions. A faster horizontal component carries the object farther during that time; a larger upward component makes it rise longer before descending. Mass does not appear in this ideal model.

We use constant gravity near Earth’s surface and a level ground plane. The launch angle ranges from 0∘0^\circ to 90∘90^\circ, and the starting height is nonnegative. Air drag, wind, spin, and Earth’s curvature are outside this model.

Explore the trajectory

Change the launch conditions, then move through the flight to inspect the position and velocity.

Projectile trajectory and velocityA projectile launched at 45 degrees with a speed of 20 meters per second from 0 meters above the ground. Dots mark equal time intervals. The highlighted projectile is at 50 percent of its flight. Velocity arrows show its current direction and components.HeightHorizontal distanceApexLaunchLanding

Dots are equally spaced in time. Solid blue: total velocity. Long dashes: horizontal velocity. Dots on the vertical arrow: vertical velocity. The ring marks the apex.

Both axes share the same distance scale, which adjusts to fit each trajectory. Velocity arrows use a separate scale.

Launch speed: 20 m s−120\,\mathrm{m\,s^{-1}}

Launch angle: 45∘45^\circ

Launch height: 0 m0\,\mathrm{m}

Inspect flight: t=1.44 st=1.44\,\mathrm{s}

Constant gravity g=9.81 m s−2g=9.81\,\mathrm{m\,s^{-2}}. Air resistance is ignored. Landing is at ground level. A horizontal launch from ground level has zero flight time in this model.

Flight time
2.88 s2.88\,\mathrm{s}
Horizontal range
40.77 m40.77\,\mathrm{m}
Maximum height above ground
10.19 m10.19\,\mathrm{m}
Current horizontal position
20.39 m20.39\,\mathrm{m}
Current height
10.19 m10.19\,\mathrm{m}
Current vertical velocity
−0.00 m s−1-0.00\,\mathrm{m\,s^{-1}}

Equations behind the flight

Open each explanation for the symbols, units, and meaning of the equation.

Split the launch velocity

vx=v0cos⁡θ,vy(0)=v0sin⁡θv_x=v_0\cos\theta,\qquad v_y(0)=v_0\sin\theta

Find the position at any time

x(t)=v0cos⁡θ ty(t)=h+v0sin⁡θ t−12gt2\begin{aligned}x(t)&=v_0\cos\theta\,t\\y(t)&=h+v_0\sin\theta\,t-\tfrac12gt^2\end{aligned}

Follow the velocity

vx(t)=v0cos⁡θ,vy(t)=v0sin⁡θ−gtv_x(t)=v_0\cos\theta,\qquad v_y(t)=v_0\sin\theta-gt

See why the path is a parabola

y=h+xtan⁡θ−gx22v02cos⁡2θy=h+x\tan\theta-\frac{gx^2}{2v_0^2\cos^2\theta}

Find the highest point

tpeak=v0sin⁡θg,H=h+v02sin⁡2θ2gt_{\mathrm{peak}}=\frac{v_0\sin\theta}{g},\qquad H=h+\frac{v_0^2\sin^2\theta}{2g}

Find the landing time and range

T=v0sin⁡θ+v02sin⁡2θ+2ghgR=v0cos⁡θ T\begin{aligned}T&=\frac{v_0\sin\theta+\sqrt{v_0^2\sin^2\theta+2gh}}{g}\\R&=v_0\cos\theta\,T\end{aligned}

When launch and landing heights match

For a launch from ground level, h=0h=0, the flight time and range simplify to

T=2v0sin⁡θg,R=v02sin⁡(2θ)gT=\frac{2v_0\sin\theta}{g},\qquad R=\frac{v_0^2\sin(2\theta)}{g}

At a fixed launch speed, the greatest range occurs at 45∘45^\circ. Complementary angles, such as 30∘30^\circ and 60∘60^\circ, give the same range but different heights and flight times. Try both in the lab with launch height set to zero.

An elevated launch has extra falling time, so these shortcuts no longer apply. Its best range angle is below 45∘45^\circ in this drag-free model. Use the general landing-time equation above when the starting height changes.

Work through a launch

A ball leaves ground level at a speed of 20 meters per second and an angle of 30 degrees.

  1. Resolve the launch velocity.
    vx=20cos⁡30∘≈17.32 m s−1,vy(0)=20sin⁡30∘=10 m s−1v_x=20\cos30^\circ\approx17.32\,\mathrm{m\,s^{-1}},\qquad v_y(0)=20\sin30^\circ=10\,\mathrm{m\,s^{-1}}
  2. Find the time to land, using g=9.81 m s−2g=9.81\,\mathrm{m\,s^{-2}}.
    T=2(10)9.81≈2.04 sT=\frac{2(10)}{9.81}\approx2.04\,\mathrm{s}
  3. Multiply the horizontal velocity by flight time.
    R=(20)2sin⁡60∘9.81≈35.31 mR=\frac{(20)^2\sin60^\circ}{9.81}\approx35.31\,\mathrm{m}
  4. Calculate the peak height.
    H=(10)22(9.81)≈5.10 mH=\frac{(10)^2}{2(9.81)}\approx5.10\,\mathrm{m}

Set these launch conditions in the lab to check the results.

What to watch for

  • At the apex, vy=0v_y=0, but gravity still acts. The object keeps moving horizontally unless the launch was straight up.
  • Time to rise equals time to fall only when the launch and landing heights match. An elevated launch takes longer to fall to the ground.
  • Doubling the launch speed at the same angle quadruples the range for equal launch and landing heights; it doubles the flight time.
  • A straight-up launch has zero horizontal range. A horizontal launch needs a positive starting height to have a flight above the ground.

Connect this to motion diagrams and the essential physics equations.

Source

Try it yourself

Predict the result first, then open the solution or check it in the lab.

  • Launch horizontally from a platform

    A ball starts 20 m20\,\mathrm{m} above the ground with horizontal speed 10 m s−110\,\mathrm{m\,s^{-1}}. Find its flight time and range.

  • Compare two launch angles

    Launch two balls from ground level at 20 m s−120\,\mathrm{m\,s^{-1}}, one at 30∘30^\circ and one at 60∘60^\circ. Which goes farther, and which stays airborne longer?